Determine the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
f(f(f(x)))+f(f(y))=f(y)+x
$$
Let's show that the function is surjective. Let $t \in \mathbb{R}$, we choose $x$ such that $x - f(0) - f(f(0)) = t$ and $u = f(f(x))$. Then

$$
f(u) = f(f(f(x))) = x - f(0) - f(f(0)) = t
$$

so $f$ is indeed surjective. We can then set $Y = f(y)$ and $x = 0$ and we get

$$
f(Y) = Y - f(f(f(0)))
$$

for all $Y \in \mathbb{R}$. Therefore, $f$ is affine of the form $f(x) = x + c$. We return to the statement

$$
f(f(f(x))) + f(f(y)) = x + 3c + y + 2c = x + y + 5c = y + c + x
$$

so $c = 0$. We verify that $f(x) = x$ is indeed a solution.