Determine all functions $ f : \mathbb R \to \mathbb R $ such that $$ f \bigl ( x - f ( y ) \bigr ) = f ( - x ) + \bigl ( f ( y ) - 2 x \bigr ) \cdot f ( - y ) $$ for all $ x , y \in \mathbb R $.
It's easy to see that $ f ( x ) = x ^ 2 $ is a function satisfying the above equation. Thus I thought it would be wise to first prove that $ f $ is an even function. The best I did is to conclude that $ f \bigl ( - f ( y ) \bigr ) = f \bigl ( - f ( - y ) \bigr ) $. Then I tried to prove that $ f ( 0 ) = 0 $ but failed.
This is a loose derivation.
Let $x = 0$, to have: $$ f(-f(y))=f(0)+f(y)\cdot f(-y) $$ Let $ y = -y$: $$ f(-f(-y))=f(0)+f(-y)\cdot f(y) $$ So $f(-(f(y)) = f(-f(-y))$, I think this is sufficient to conclude that $f$ is even, by apply $f^{-1}$ on both sides and multiply $-1$.
Now with $f$ even, $$ f(-f(y))=f(0)+f(y)\cdot f(y) $$ Let $f(y) = x$, we have: $$ f(-x)=f(x)=f(0)+x^2 $$
You can not determine what $f(0)$ is.