Denote by $\mathbb{R}^{+}$ the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that
$$x f\left(x^{2}\right) f(f(y))+f(y f(x))=f(x y)\left(f\left(f\left(x^{2}\right)\right)+f\left(f\left(y^{2}\right)\right)\right)$$
for all positive real numbers $x$ and $y$. Answer. $f(x)=\frac{1}{x}$ for any $x \in \mathbb{R}^{+}$. ! hence $f(1)=1$. Swapping $x$ and $y$ in (1) and comparing with (1) again, we find
$$x f\left(x^{2}\right) f(f(y))+f(y f(x))=y f\left(y^{2}\right) f(f(x))+f(x f(y)) .$$
Taking $y=1$ in (2), we have $x f\left(x^{2}\right)+f(f(x))=f(f(x))+f(x)$, that is,
$$f\left(x^{2}\right)=\frac{f(x)}{x}$$
Take $y=1$ in (1) and apply (3) to $x f\left(x^{2}\right)$. We get $f(x)+f(f(x))=f(x)\left(f\left(f\left(x^{2}\right)\right)+1\right)$, which implies
$$f\left(f\left(x^{2}\right)\right)=\frac{f(f(x))}{f(x)}$$
For any $x \in \mathbb{R}^{+}$, we find that
$$f\left(f(x)^{2}\right) \stackrel{(3)}{=} \frac{f(f(x))}{f(x)} \stackrel{(4)}{=} f\left(f\left(x^{2}\right)\right) \stackrel{(3)}{=} f\left(\frac{f(x)}{x}\right)$$
It remains to show the following key step.
- Claim. The function $f$ is injective.
Proof. Using (3) and (4), we rewrite (1) as
$$f(x) f(f(y))+f(y f(x))=f(x y)\left(\frac{f(f(x))}{f(x)}+\frac{f(f(y))}{f(y)}\right) .$$
Take $x=y$ in (6) and apply (3). This gives $f(x) f(f(x))+f(x f(x))=2 \frac{f(f(x))}{x}$, which means
$$f(x f(x))=f(f(x))\left(\frac{2}{x}-f(x)\right)$$
Using (3), equation (2) can be rewritten as
$$f(x) f(f(y))+f(y f(x))=f(y) f(f(x))+f(x f(y))$$
Suppose $f(x)=f(y)$ for some $x, y \in \mathbb{R}^{+}$. Then (8) implies
$$f(y f(y))=f(y f(x))=f(x f(y))=f(x f(x))$$
Using (7), this gives
$$f(f(y))\left(\frac{2}{y}-f(y)\right)=f(f(x))\left(\frac{2}{x}-f(x)\right)$$
Noting $f(x)=f(y)$, we find $x=y$. This establishes the injectivity.
By the Claim and (5), we get the only possible solution $f(x)=\frac{1}{x}$. It suffices to check that this is a solution. Indeed, the left-hand side of (1) becomes
$$x \cdot \frac{1}{x^{2}} \cdot y+\frac{x}{y}=\frac{y}{x}+\frac{x}{y},$$
while the right-hand side becomes
$$\frac{1}{x y}\left(x^{2}+y^{2}\right)=\frac{x}{y}+\frac{y}{x} .$$
The two sides agree with each other.