Baltic Way 2019

Putting $y=-1$ and $x=0$, we obtain $f(-1)=0$. Suppose that $f$ has another zero, i.e., there is an $a\ne -1$ with $f(a)=0$. Putting $x=0$ and $y=-a$ gives $f(-a^2)=0$. Now choosing $y=a$, we obtain $f(x-a)=0$. Hence, $f(x)=0$ for all $x$ which clearly is a solution.

Let us assume now that $x=-1$ is the only zero of $f$. Then $x=y-1$ gives $f((y-1)f(y)-y^2)=0$. Consequently, we have $(y-1)f(y)=y^2-1$ for all $y$ which implies $f(x)=x+1$ for all $x\ne 1$. Setting $y=2,x=3$ gives $f(1)=2$. Hence, $f(x)=x+1$ for all $x$ which is another solution.

Altogether, the only solutions are $f(x)=0$ and $f(x)=x+1$.