Croatian Mathematical Olympiad

Plugging $(x, y) \leftarrow (f(x), y)$ into the given equation, it follows that
$$f(f(x)^2 f(y)^2) = f(f(x))^2 f(y)$$
holds for all positive rational numbers $x$ and $y$, hence $f(f(y))^2 f(x) = f(f(x)^2 f(y)^2) = f(f(x))^2 f(y)$, i.e.
$$\frac{f(x)}{f(y)} = \left( \frac{f(f(x))}{f(f(y))} \right)^2 = \left( \frac{f^2(x)}{f^2(y)} \right)^2,$$
where $f^k$ denotes the $k$th iterate of $f$.
By mathematical induction, we can show that
$$\frac{f(x)}{f(y)} = \left( \frac{f^n(x)}{f^n(y)} \right)^{2^{n-1}}$$
holds, meaning that $f(x)/f(y)$ is the $(2^n)^{th}$ power of a rational number for all positive integers $n$. Hence, $f(x) = f(y)$ holds for all positive rational numbers $x$ and $y$, i.e. $f(x)$ is a constant.
Finally, from the given equation we get that $f(x) = 1$ for all positive rational numbers $x$ is the only solution.