If $f(x-f(y))=f(-x)+(f(y)-2x)\cdot f(-y)$ what is $f(x)$ 
Determine all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(x-f(y))=f(-x)+(f(y)-2x)\cdot f(-y), \quad \forall x,y \in \mathbb{R}$$

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 $f(-f(y))=f(-f(-y))$. 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.
