Find the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x, y \in \mathbb{R}$ we have $f(x-f(x-y))+x=$ $f(x+y)$.
Let $f$ be a solution. We set $y=-x$ which allows us to have

$$
f(x-f(2 x))=f(0)-x
$$

From this, we deduce that the function $f$ is surjective. From there, we fix $y \in \mathbb{R}$ and we take an $x \in \mathbb{R}$ such that $-f(x-y)=y$ whose existence is guaranteed by the surjectivity. We then have $f(x+y)=f(x-f(x-y))$ but we also have $f(x-f(x-y))+x=f(x+y)$, so $x=0$. This can be rewritten as $f(-y)=-y$. Conversely, the identity function is a solution to the problem because $x-(x-y)+x=x+y$.