Bulgarian Mathematical Competitions

Solution:
Now let $x \leq y \leq z$, $x, y, z \in (0,1)$. Then using (a), we get that $f\left(x y^{k-1}, z\right) = f\left(x, y z^{k-1}\right)$. Hence it follows from the above that
$$\{x y^{k-1} z^{k-1}, x^{k-1} y^{(k-1)^2} z\} \cap \{x y^{k-1} z^{(k-1)^2}, x^{k-1} y z^{k-1}\} \neq \varnothing$$
This easily implies that either $k=1$ and $f(x, y) = \min\{x, y\}$ or $k=2$ and $f(x, y) = x y$. A direct verification shows that both functions are solutions for the respective $k$.