55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour)

Substitute $y=1$ and $y=-1$: $$f(f(x)-1)=f(x^2)+f(1)-2f(x)\text{ and }f(f(x)-1)=f(x^2)+f(-1)-2f(-x).(\ast )$$ Combining both equalities, get $f(1)-2f(x)=f(-1)-2f(-x)$. Hence, if $x=1$: $f(1)=f(-1)$, so $f(x)=f(-x)$, the function $f$ is even.

If $x=y=1$, $f(f(1)-1)=0$, in other words there exists such number $b$ that $f(b)=0$. If $x=b$ in $(\ast )$ then $$f(f(b)-1)=f(b^2)+f(1),f(-1)=f(b^2)+f(1),\text{so }f(b^2)=0.$$ Let substitute $x=b$ and $y=0$ in the initial equality: $f(f(b))=f(b^2)-2f(0)$, so $3f(0)=f(b^2)=0$. If $x=0$ then $$f(y^2)=y^{2}f(y).(\ast \ast )$$ There are two cases. Case 1: there exists $b\in \mathbb{R}$, $b\ne 0$ such that $f(b)=0$. As is shown above, $f(b^2)=0$. Let substitute $x=b$ in the initial equality: