If $x\leq y$, we have $f(x, y)=f\left(\frac{x}{y}, 1\right)\cdot y^k=xy^{k-1}$ by the third and fourth rules. If $y\leq x$, then we have $f(x, y)=f(y, x)=yx^{k-1}$. Now assume $x\geq y$ and choose small enough nonzero $z$. Then $f(f(x, y), z)=f(x^{k-1}y, z)=(x^{k-1}y)^{k-1}z$. Also $f(x, f(y, z))=f(x, y^{k-1}z)=x^{k-1}y^{k-1}z$ so that $(k-1)^2=(k-1)$. Hence $k=1$ or $k=2$. If $k=1$, then $f(x, y)=\min(x, y)$ and if $k=2$, then $f(x, y)=xy$, both of which work.

tenniskidperson3 wrote: If $x\leq y$, we have $f(x, y)=f\left(\frac{x}{y}, 1\right)\cdot y^k=xy^{k-1}$ by the third and fourth rules. This is not the right method because you can't set $ z=\frac{1}{y}\ge 1. $ This can be obtained by setting $ y=1 $ and then you get $ f(zx,z)=z^kx. $ Now if you set $ zx=y $, you get $ f(y,z)=yz^{k-1} $ for $ y\le z. $

Fortunately, I didn't set $z=\frac{1}{y}$ but instead $z=y$; and the $x$ in the rule is our $\frac{x}{y}\leq 1$; and the $y$ in the rule is just $1$.

$P(x,x)$ $\implies$ $f(x,x)=f(x \cdot 1, x \cdot 1)=x^kf(1,1)$ but $f(1,1)=1$ (from $2$) $\implies$ $f(x,x)=x^k$ ($*$). $P(f(x,x),f(x,x))$ $\implies$ $f(f(x,x),f(x,x))=x^{k^2}$ (from $*$). $f(f(x,x),f(x,x))=f(f(f(x,x),x),x)=f(f(x^k,x),x)=f(x^k \cdot f(x^{k-1},1), x)=f(x^k \cdot x^{k-1},x)=f(x^{2k-1},x)=x^kf(x^{2k-2},1)=x^{3k-2}$ hence $k^2=3k-2$ $\implies$ $k=1,2$. For $k=2$ take $f(x,y)=xy$ and for $k=1$ take $f(x,y)=min \left\{x,y \right\}$. So answer is $k=1,2$.