This is almost the same problem; the only time we used commutativity is when we established the case $x\geq y$, which we can do here by just using the same method twice. So we must have $k=1$ or $k=2$, and $f(x, y)=\min(x, y)$ for $k=1$ and $xy$ for $k=2$ are the only solutions.

Assume that: $y \ge x$: Thus: $f(x, y) = y^k f(x/y, 1) = x y^{k-1}$. Thus: $f(x, y) = \max(x, y)^{k-1} \min(x, y)$. It clearly satisfies the ii), iii) conditions. Plugging into condition i) $y=z$ with $y>x$: $f(f(x, y), y) = f(x y^{k-1}, y)$. Letting $x < \frac{1}{y}^{2-k}$, then $f(x y^{k-1}, y)=y^{2k-2}x$. Now, $f(x, f(y, y)) = f(x, y^k) = y^{k(k-1)}x$ where we let $x < y^k$. Thus: $2k-2=k^2-k$ or $k=1, 2$. Plugging it back clearly works and we are done with: $f(x, y) = \min(x, y), xy$.