Determine all functions $f$ from $\mathbb{R}$ to $\mathbb{R}$ such that, for all real $x$ and $y$, the following equality holds:

$$
f(x+y)=f(x-y)+f(f(1-x y))
$$
For $y=0$ we find that $f(x)=f(x)+f(f(1))$ so $f(f(1))=0$. For $x=0$ we find $f(y)=f(-y)+f(f(1))$ so $f(y)=f(-y)$ for all $y$.
For $y=1$ we get $f(x+1)=f(x-1)+f(f(1-x))$. By replacing $x$ with $-x+2$, we get $f(-x+3)=f(-x+1)+f(f(x-1))$.

Since $f(x-1)=f(1-x)$, then $f(f(1-x))=f(f(x-1))$. Consequently, $f(x+1)=f(x-1)+f(-x+3)-f(-x+1)=f(x-3)$ because $f(-t)=f(t)$ for all $t$.
We deduce, by replacing $x$ with $x+1$, that $f(x+2)=f(x-2)$ for all $x$.
Take $y=2$ in the functional equation. We have $f(x+2)=f(x-2)+f(f(1-2x))$, so $f(f(1-2x))=0$ for all $x$. By replacing $x$ with $(1-t)/2$, we find that $f(f(t))=0$ for all $t$.
We return to the functional equation: $f(x+y)=f(x-y)+f(f(1-xy))$, so $f(x+y)=f(x-y)$ for all $x$ and $y$. We take $x=y=t/2$: we get $f(t)=f(0)$, so $f$ is constant. Since $f(f(1))=0$, this constant is zero, and finally $f(x)=0$ for all $x$.