Japan Mathematical Olympiad

If we substitute $(0,0)$ for $(x,y)$ into the given identity, we get $f((f(0){)}^2)=0$. And applying this fact after substituting $(0,(f(0){)}^2)$ for $(x,y)$ in the given identity, we obtain the fact that $f(0)=0$ must hold.

Let $t$ be an arbitrary non-zero real number, and substitute $(t,t)$ for $(x,y)$ in the given identity, we get $0=t^2-tf(t)$, from which we conclude that $f(t)=t$ holds. Combining with the fact that $f(0)=0$ holds as well, we conclude that $f(t)=t$ holds for any real number $t$.

Clearly, the function $f(t)=t$ satisfies the given identity for any pair of real numbers $(x,y)$, so we conclude that the function $f(t)=t$ is the only solution for the problem.