We first show that $f$ is injective. Suppose that $f(y) = f(z)$. Then we have:
$$\frac{xy}{x+y} = f(f(x) + f(y)) = f(f(x) + f(z)) = \frac{xz}{x+z}$$
Taking reciprocals of the first and last expression implies $y = z$, as required for injectivity.

Note that the right hand side of the functional equation can be written as
$$\frac{xy}{x+y} = \frac{1}{\frac{1}{x} + \frac{1}{y}}$$
If $x = \frac{k}{n-a}$ and $y = \frac{k}{n+a}$ with $a \ne \pm n$, the expression $\frac{1}{x} + \frac{1}{y} = \frac{2n}{k}$ does not depend on $a$. Hence, for such $x$ and $y$, the right hand side of the functional equation has a value that does not depend on $a$. In particular, with $n \ge 2$, $a = 0$ and $a = 1$, the functional equation implies
$$f\left(f\left(\frac{k}{n}\right) + f\left(\frac{k}{n}\right)\right) = f\left(f\left(\frac{k}{n-1}\right) + f\left(\frac{k}{n+1}\right)\right).$$
As $f$ is injective, we can equate the arguments of the two outer $f$'s to deduce:
$$f\left(\frac{k}{n+1}\right) - f\left(\frac{k}{n}\right) = f\left(\frac{k}{n}\right) - f\left(\frac{k}{n-1}\right). \quad (20)$$

Next, we are going to prove that there are rational numbers $a, b$ such that
$$f(x) = a + \frac{b}{x} \quad \text{for all } x \in \mathbb{Q}_+. \qquad (21)$$

Our first proof of (21) starts with the claim that for each $k \in \mathbb{Q}_+$ and any integer $n \ge 1$ there exist rational numbers $a(k)$ and $b(k)$ such that
$$f\left(\frac{k}{n}\right) = a(k) + b(k)\frac{n}{k}. \qquad (22)$$
Indeed, from (20) we know that
$$b(k) := k \left( f\left(\frac{k}{n+1}\right) - f\left(\frac{k}{n}\right) \right)$$
does not depend on $n$. In particular, we have $\frac{b(k)}{k} = f(\frac{k}{2}) - f(k)$. If we let $a(k) = 2f(k) - f(\frac{k}{2})$ we obtain $f(k) = a(k) + \frac{b(k)}{k}$. This is the case $n = 1$ of (20) which gets our proof by induction going. For the inductive step, we use the definition of $b(k)$ and the inductive hypothesis:
$$f\left(\frac{k}{n+1}\right) = f\left(\frac{k}{n}\right) + \frac{b(k)}{k} = a(k) + b(k)\frac{n}{k} + \frac{b(k)}{k} = a(k) + b(k)\frac{n+1}{k}.$$
We next show that $a(k)$ and $b(k)$ do not depend on $k$. For any $k \in \mathbb{Q}_+$ and $n \in \mathbb{Z}_+$ we have
$$f\left(\frac{k}{n}\right) = f\left(\frac{k'}{1}\right) \quad \text{and} \quad f\left(\frac{k}{2n}\right) = f\left(\frac{k'}{2}\right) \quad \text{where} \quad k' = \frac{k}{n}$$
and (22) gives us
$$a(k) + b(k) \frac{n}{k} = a(k') + b(k') \frac{n}{k} \quad \text{and} \quad a(k) + b(k) \frac{2n}{k} = a(k') + b(k') \frac{2n}{k}.$$
Subtracting the second from twice the first equation, and the first from the second, gives
$$a(k) = a(k') \quad \text{and} \quad b(k) \frac{n}{k} = b(k') \frac{n}{k'} \quad \text{hence}$$
$$a(k) = a\left(\frac{k}{n}\right) \quad \text{and} \quad b(k) = b\left(\frac{k}{n}\right) \quad \text{for all } k \in \mathbb{Q}_+, n \in \mathbb{Z}_+.$$
Letting $k = 1$ we now obtain $a(1) = a(\frac{1}{n})$ for all $n \in \mathbb{Z}_+$. Using $k = n/m$ with arbitrary $m, n \in \mathbb{Z}_+$ we obtain $a(\frac{n}{m}) = a(\frac{1}{m})$. Together these show that $a(k) = a(1)$ for all $k \in \mathbb{Q}_+$. Similarly it follows that $b(k) = b(1)$ for all $k \in \mathbb{Q}_+$. This establishes (21) with $a = a(1)$ and $b = b(1)$.

For our second proof of (21) we define
$$g(k) := f\left(\frac{k}{n+1}\right) - f\left(\frac{k}{n}\right)$$
For all $k \in \mathbb{Q}_+$ and integers $n \ge 1$. From (20) we know that the right hand side does not depend on $n$. A straightforward induction yields
$$f\left(\frac{k}{n+i}\right) = f\left(\frac{k}{n}\right) + ig(k) \quad \text{for all } k \in \mathbb{Q}_+, n \ge 1, i \ge 0. \quad (23)$$
Next we show that $vg(1) = ug(u/v)$ for all positive integers $u, v$. To see this, we first substitute $k = 1$ and $i = n = v$ in equation (23) to obtain
$$f\left(\frac{1}{2v}\right) = f\left(\frac{1}{v}\right) + vg(1).$$
Next we substitute $k = u/v$ and $i = n = u$ in equation (23) and obtain
$$f\left(\frac{1}{2v}\right) = f\left(\frac{1}{v}\right) + ug\left(\frac{u}{v}\right).$$
Comparing these two equations we get the desired equality, which means that $kg(k) = g(1)$ for all $k \in \mathbb{Q}_+$.
Note that substituting $n = 1$ and $i = m - 1 \ge 0$ in (23) gives
$$f\left(\frac{k}{m}\right) = f(k) + (m-1)g(k). \quad (24)$$
For $k \in \mathbb{Q}_+$ we now define $a(k) = f(k) - g(k)$. Setting $k = 1$ in (24) gives
$$f\left(\frac{1}{m}\right) = f(1) + (m-1)g(1) = a(1) + mg(1).$$
With $k = n/m$ and $m$ replaced by $n$ in (24) we obtain
$$\begin{aligned} f\left(\frac{1}{m}\right) &= f\left(\frac{\frac{n}{m}}{n}\right) = f\left(\frac{n}{m}\right) + (n-1)g\left(\frac{n}{m}\right) \\ &= a\left(\frac{n}{m}\right) + ng\left(\frac{n}{m}\right) = a\left(\frac{n}{m}\right) + mg(1). \end{aligned}$$
Comparing these two equations, we obtain $a(n/m) = a(1)$ for all positive integers $m, n$. If we now let $a = a(1)$ and $b = g(1)$ we finally obtain
$$f(k) = a(k) + g(k) = a(1) + \frac{g(1)}{k} = a + \frac{b}{k}$$
for all $k \in \mathbb{Q}_+$. This finishes the second proof of (21).

Our penultimate step is to show that $a = 0$ in (21). To show this, we start by substituting (21) into the original functional equation with $y = x$:
$$\frac{x}{2} = f(f(x) + f(x)) = f(2f(x)) = f\left(2a + \frac{2b}{x}\right) = a + \frac{bx}{2ax + 2b}.$$
This implies $x(ax+b) = 2a(ax+b)+bx$, i.e. $a(2ax-x^2+2b) = 0$. So, either $a=0$ or $2b = x^2-2ax$ for all $x \in \mathbb{Q}_+$. But the polynomial $x^2-2ax$ is not constant, so we must have $a=0$. Thus, $f(x) = \frac{b}{x}$ with $b \in \mathbb{Q}_+$. An easy check reveals that all such functions satisfy the functional equation. As $f(1) = 2023$, we must have $b = 2023$ so that $f(2023) = 1$.