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.