Ukrainian National Mathematical Olympiad - Third Round (Second Tour)

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