CroatianCompetitions

Taking $x = 0$ in the given equation we get $f(f(y)) = y$ for every $y \in \mathbb{R}$.
Therefore, we have
$$f(y - x^2) = f(f(x^2 + f(y))) = x^2 + f(y). \qquad (3.1)$$
for every $x, y \in \mathbb{R}$.
Taking $y = x^2$ in (3.1) we get $f(0) = x^2 + f(x^2)$, that is
$$f(x^2) = -x^2 + f(0),$$
while taking $y = 0$ in (*) gives us
$$f(-x^2) = x^2 + f(0).$$
Let $c = f(0)$. From the above we conclude that
$$f(x) = -x + c, \quad \forall x \in \mathbb{R}.$$
It is easy to verify that all the functions of the form $f(x) = -x + c$ for $c \in \mathbb{R}$ satisfy the given equation:
$$f(x^2 + f(y)) = f(x^2 - y + c) = -(x^2 - y + c) + c = -x^2 + y.$$