AustriaMO2013

Putting $x=y=1$ in (2) yields $f(1)=f(1{)}^2$, that is $f(1)\in \{0,1\}$.

Case 1: $f(1)=0$ (a) If $k=0$, (2) becomes for $x=y=t$, $t\ne 0$: $f(1)=t^{2}f(t{)}^2$, whence $f(t)=0$, $t\ne 0$. Let $k\ne 0$. Then $y=1$ and $x=t$ in (2) yields $f(t^k)=tf(t)f(1)=0$. Letting $x=y$ in (2) finally leads to $f((x^2{)}^k)=x^{2}f(x{)}^2$, that is $0=x^{2}f(x{)}^2$, whence $f(x)=0$, $x\ne 0$.

Case 2: $f(1)=1$ Letting $x=t$ and $y=1/t$, $t\ne 0$, in (2) shows $f(1)=f(t)\cdot f(1/t)$. Therefore, $f(t)\ne 0$, $t\ne 0$. Putting $x=y=-1$ in (2) yields $f((-1{)}^{2k})=(-1{)}^{2}f(-1{)}^2$, that is $f(-1{)}^2=1$. Thus $f(-1)\in \{-1,1\}$. (a) Let $k$ be odd. Then $x=1$ and $y=-1$ in (2) leads to $$f((-1{)}^k)=-f(-1)(3)$$ that is $f(-1)=-f(-1)$. Therefore, we get the contradiction $f(-1)=0$. (b) For $k$ even we get from (3): $f(-1)=-1$.

i) If $k=0$, (2) yields for $y=1$ and $x=t$, $t\ne 0$: $f(1)=tf(t)$ that is $f(t)=1/t$.

ii) For $k\ne 0$ we get from (2) for $x=1$, $y=t$, $t\ne 0$: $$f(t^k)=tf(t).(4)$$ This and (2) imply $$xyf(x)f(y)=f(x^k{y}^k)=f((xy{)}^k)=xyf(xy).$$ Therefore, $$f(xy)=f(x)f(y)(5)$$ Putting $x=t$ and $y=t^{k-1}$, $t\ne 0$, yields in view of (4) and (5): $$tf(t)=f(t^k)=f(t\cdot t^{k-1})=f(t)f(t^{k-1}).$$ Therefore, $f(t^{k-1})=t$, $t\ne 0$. As $k$ is even all real numbers $x$, $x\ne 0$, have a unique representation $x=t^{k-1}$, $t\ne 0$. Thus, finally $f(x)=x^{1/(k-1)}=t^{k-1/2}$.

It is easily checked that all functions obtained in the course of our solution satisfy (2).

Summary: All solutions to (2) are given by $f(x)=0$, $x\in \mathbb{R}$, for all integers $k$ and additionally $f(x)=$, if $k$ is an even integer.