For $y = 0$ we have $f(f(x)) = x + (f(0))^n$, for any $x \in \mathbb{R}$, implying that $f$ is bijective. Let $a$ such that $f(a) = 0$. For $y = a$ we deduce $f(f(x)) = x$, for all real $x$, so $f(0) = 0$.

For $x = 0$ we get $f(y f(y)) = f(y)^n$, for real $y$. Substituting $y$ with $f(y)$, we get $f(y)^n = y^n$.

i) For odd $n$ we get $f(y) = y$, which gives a contradiction in the given relation.

ii) For even $n$ we obtain $f(y) \in \{-y, y\}$, for any $y \in \mathbb{R}$. Thus $y^n \in \{-y^2, y^2\}$ for all real $y$, which simply imply $n = 2$. Thus $f(f(x) + y f(y)) = x + y^2$, for $x, y \in \mathbb{R}$.

It is easy to check that the identity function and minus the identity function, verify the relation. We shall show that they are the only ones. If not, if there are $a, b \neq 0$ such that $f(a) = a$ and $f(b) = -b$, than for $x = a, y = b$ we should have $f(a - b^2) = a + b^2 \in \{a - b^2, -a + b^2\}$, that is $a = 0$ or $b = 0$, a contradiction.