Slovenian Mathematical Olympiad, 2016

Let us show that the function $f$ is surjective. Substituting $y \mapsto \frac{(f(x))^2}{y f(f(x))}$ in the initial equation we get
$$f\left(\frac{x}{f\left(\frac{(f(x))^2}{y f(f(x))}\right)}\right) = y,$$
which means that for any $y$ there exists a number which is mapped into $y$ by $f$. Hence, $f$ is surjective.

Surjectivity implies the existence of $c \in \mathbb{R}^+$, such that $f(c) = 1$. Insert $y = c$ into the initial equation.
$$\begin{aligned} & f\left(\frac{x}{f(c)}\right) = \frac{(f(x))^2}{c f(f(x))} \\ \Rightarrow \quad & f(x) = \frac{(f(x))^2}{c f(f(x))} \\ \Rightarrow \quad & c f(f(x)) = f(x), \end{aligned}$$
but because $f$ is surjective we can substitute $f(x)$ for any $y \in \mathbb{R}^+$. This implies that
$$f(y) = \frac{1}{c} y \quad \text{for all } y \in \mathbb{R}^{+}.$$
It is easy to check that all functions of the form $f(x) = kx$, where $k \in \mathbb{R}^+$ is an arbitrary constant, satisfy the original equation.