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

$$
f(f(f(x)))+f(f(y))=f(y)+x
$$
First, we notice that by taking $y=0$, we obtain:

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

thus $f$ is surjective. Let $a$ and $b$ be such that $f(a)=f(b)$, by replacing $x$ with $a$, we have

$$
f(y)+a=f(f(f(a)))+f(f(y))=f(f(f(b)))+f(f(y))=f(y)+b
$$

thus $a=b$, so $f$ is injective. Since $f$ is surjective, we can replace $f(y)$ with $t$ which can take any value in $\mathbb{R}$. By setting $x=0$, we then have:

$$
f(f(f(0)))+f(t)=t
$$

thus $f$ is of the form $f(x)=x-a$. By substituting into the initial equation, we find that the only solution is $f(x)=x$