First we prove that the function $f(x)=1 / x$ satisfies the condition of the problem statement. The AM-GM inequality gives
$$\frac{x}{y}+\frac{y}{x} \geqslant 2$$
for every $x, y>0$, with equality if and only if $x=y$. This means that, for every $x>0$, there exists a unique $y>0$ such that
$$\frac{x}{y}+\frac{y}{x} \leqslant 2$$
namely $y=x$.

Let now $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ be a function that satisfies the condition of the problem statement. We say that a pair of positive real numbers $(x, y)$ is good if $x f(y)+y f(x) \leqslant 2$. Observe that if $(x, y)$ is good, then so is $(y, x)$.

Lemma 1.0. If $(x, y)$ is good, then $x=y$.

Proof. Assume that there exist positive real numbers $x \neq y$ such that $(x, y)$ is good. The uniqueness assumption says that $y$ is the unique positive real number such that $(x, y)$ is good. In particular, $(x, x)$ is not a good pair. This means that
$$x f(x)+x f(x)>2$$
and thus $x f(x)>1$. Similarly, $(y, x)$ is a good pair, so $(y, y)$ is not a good pair, which implies $y f(y)>1$. We apply the AM-GM inequality to obtain
$$x f(y)+y f(x) \geqslant 2 \sqrt{x f(y) \cdot y f(x)}=2 \sqrt{x f(x) \cdot y f(y)}>2 .$$
This is a contradiction, since $(x, y)$ is a good pair.

By assumption, for any $x>0$, there always exists a good pair containing $x$, however Lemma 1 implies that the only good pair that can contain $x$ is $(x, x)$, so
$$x f(x) \leqslant 1 \quad \Longleftrightarrow \quad f(x) \leqslant \frac{1}{x},$$
for every $x>0$.

Solution 1.1. We give an alternative way to prove that $f(x)=1 / x$ assuming $f(x) \leqslant 1 / x$ for every $x>0$.
Indeed, if $f(x)<1 / x$ then for every $a>0$ with $f(x)<1 / a<1 / x$ (and there are at least two of them), we have
$$a f(x)+x f(a)<1+\frac{x}{a}<2 .$$
Hence $(x, a)$ is a good pair for every such $a$, a contradiction. We conclude that $f(x)=1 / x$.

Solution 1.2. We can also conclude from Lemma 1 and $f(x) \leqslant 1 / x$ as follows.
Lemma 2. The function $f$ is decreasing.
Proof. Let $y>x>0$. Lemma 1 says that $(x, y)$ is not a good pair, but $(y, y)$ is. Hence
$$x f(y)+y f(x)>2 \geqslant 2 y f(y)>y f(y)+x f(y),$$
where we used $y>x$ (and $f(y)>0$ ) in the last inequality. This implies that $f(x)>f(y)$, showing that $f$ is decreasing.

We now prove that $f(x)=1 / x$ for all $x$. Fix a value of $x$ and note that for $y>x$ we must have $x f(x)+y f(x)>x f(y)+y f(x)>2$ (using that $f$ is decreasing for the first step), hence $f(x)>\frac{2}{x+y}$. The last inequality is true for every $y>x>0$. If we fix $x$ and look for the supremum of the expression $\frac{2}{x+y}$ over all $y>x$, we get
$$f(x) \geqslant \frac{2}{x+x}=\frac{1}{x} .$$
Since we already know that $f(x) \leqslant 1 / x$, we conclude that $f(x)=1 / x$.

As in the first solution, we note that $f(x)=1 / x$ is a solution, and we set out to prove that it is the only one. We write $g(x)$ for the unique positive real number such that $(x, g(x))$ is a good pair. In this solution, we prove Lemma 2 without assuming Lemma 1.

Lemma 2. The function $f$ is decreasing.
Proof. Consider $x<y$. It holds that $y f(g(y))+g(y) f(y) \leqslant 2$. Moreover, because $y$ is the only positive real number such that $(g(y), y)$ is a good pair and $x \neq y$, we have $x f(g(y))+g(y) f(x)>2$. Combining these two inequalities yields
$$x f(g(y))+g(y) f(x)>2 \geqslant y f(g(y))+g(y) f(y),$$
or $f(g(y))(x-y)>g(y)(f(y)-f(x))$. Because $g(y)$ and $f(g(y))$ are both positive while $x-y$ is negative, it follows that $f(y)<f(x)$, showing that $f$ is decreasing.

We now prove Lemma 1 using Lemma 2. Suppose that $x \neq y$ but $x f(y)+y f(x) \leqslant 2$. As in the first solution, we get $x f(x)+x f(x)>2$ and $y f(y)+y f(y)>2$, which implies $x f(x)+y f(y)>2$. Now
$$x f(x)+y f(y)>2 \geqslant x f(y)+y f(x)$$
implies $(x-y)(f(x)-f(y))>0$, which contradicts the fact that $f$ is decreasing. So $y=x$ is the unique $y$ such that $(x, y)$ is a good pair, and in particular we have $f(x) \leqslant 1 / x$.

We can now conclude the proof as in any of the Solutions 1.x.

As in the other solutions we verify that the function $f(x)=1 / x$ is a solution. We first want to prove the following lemma:

Lemma 3. For all $x \in \mathbb{R}_{>0}$ we actually have $x f(g(x))+g(x) f(x)=2$ (that is: the inequality is actually an equality).

Proof. We proceed by contradiction: Assume there exists some number $x>0$ such that for $y=g(x)$ we have $x f(y)+y f(x)<2$. Then for any $0<\epsilon<\frac{2-x f(y)-y f(x)}{2 f(x)}$ we have, by uniqueness of $y$, that $x f(y+\epsilon)+(y+\epsilon) f(x)>2$. Therefore
$$\begin{align*} f(y+\epsilon) & >\frac{2-(y+\epsilon) f(x)}{x}=\frac{2-y f(x)-\epsilon f(x)}{x} \\ & >\frac{2-y f(x)-\frac{2-x f(y)-y f(x)}{2}}{x} \\ & =\frac{2-x f(y)-y f(x)}{2 x}+f(y)>f(y) \tag{1} \end{align*}$$
Furthermore, for every such $\epsilon$ we have $g(y+\epsilon) f(y+\epsilon)+(y+\epsilon) f(g(y+\epsilon)) \leqslant 2$ and $g(y+\epsilon) f(y)+y f(g(y+\epsilon))>2$ (since $y \neq y+\epsilon=g(g(y+\epsilon)))$. This gives us the two inequalities
$$f(g(y+\epsilon)) \leqslant \frac{2-g(y+\epsilon) f(y+\epsilon)}{y+\epsilon} \quad \text{ and } \quad f(g(y+\epsilon))>\frac{2-g(y+\epsilon) f(y)}{y} .$$
Combining these two inequalities and rearranging the terms leads to the inequality
$$2 \epsilon<g(y+\epsilon)[(y+\epsilon) f(y)-y f(y+\epsilon)] .$$
Moreover combining with the inequality (1) we obtain
$2 \epsilon<g(y+\epsilon)\left[(y+\epsilon) f(y)-y \frac{2-x f(y)-y f(x)}{2 x}+f(y)\right]=g(y+\epsilon)\left[\epsilon f(y)-y \frac{2-x f(y)-y f(x)}{2 x}\right]$.
We now reach the desired contradiction, since for $\epsilon$ sufficiently small we have that the left hand side is positive while the right hand side is negative.

With this lemma it then follows that for all $x, y \in \mathbb{R}_{>0}$ we have
$$x f(y)+y f(x) \geqslant 2$$
since for $y=g(x)$ we have equality and by uniqueness for $y \neq g(x)$ the inequality is strict.

In particular for every $x \in \mathbb{R}_{>0}$ and for $y=x$ we have $2 x f(x) \geqslant 2$, or equivalently $f(x) \geqslant 1 / x$ for all $x \in \mathbb{R}_{>0}$. With this inequality we obtain for all $x \in \mathbb{R}_{>0}$
$$2 \geqslant x f(g(x))+g(x) f(x) \geqslant \frac{x}{g(x)}+\frac{g(x)}{x} \geqslant 2$$
where the first inequality comes from the problem statement. Consequently each of these inequalities must actually be an equality, and in particular we obtain $f(x)=1 / x$ for all $x \in \mathbb{R}_{>0}$.

Again, let us prove that $f(x)=1 / x$ is the only solution. Let again $g(x)$ be the unique positive real number such that $(x, g(x))$ is a good pair.

Lemma 4. The function $f$ is strictly convex.
Proof. Consider the function $q_{s}(x)=f(x)+s x$ for some real number $s$. If $f$ is not strictly convex, then there exist $u<v$ and $t \in(0,1)$ such that
$$f(t u+(1-t) v) \geqslant t f(u)+(1-t) f(v) .$$
Hence
$$\begin{aligned} q_{s}(t u+(1-t) v) & \geqslant t f(u)+(1-t) f(v)+s(t u+(1-t) v) \\ & =t q_{s}(u)+(1-t) q_{s}(v) \end{aligned}$$
Let $w=t u+(1-t) v$ and consider the case $s=f(g(w)) / g(w)$. For that particular choice of $s$, the function $q_{s}(x)$ has a unique minimum at $x=w$. However, since $q_{s}(w) \geqslant t q_{s}(u)+(1-t) q_{s}(v)$, it must hold $q_{s}(u) \leqslant q_{s}(w)$ or $q_{s}(v) \leqslant q_{s}(w)$, a contradiction.

Lemma 5. The function $f$ is continuous.
Proof. Since $f$ is strictly convex and defined on an open interval, it is also continuous.

As in Solution 1, we can prove that $f(x) \leqslant 1 / x$. If $f(x)<1 / x$, then we consider the function $h(y)=x f(y)+y f(x)$ which is continuous. Since $h(x)<2$, there exist at least two distinct $z \neq x$ such that $h(z)<2$ giving that $(x, z)$ is good pair for both values of $z$, a contradiction. We conclude that $f(x)=1 / x$ as desired.