Let us label the equation given in the problem by $(\star)$. We will show that the functions $f$ satisfying the equation $(\star)$ are given by the following:
$$f(x) = 0, \quad x^2, \quad -x^2, \quad x^2 - 1, \quad 1 - x^2.$$
It is easy to check that all of these functions satisfy the equation $(\star)$.
We will show in the sequel that there are no other function which satisfies the equation $(\star)$.
So, let a function $f$ satisfy the equation $(\star)$ and suppose for some $t$, $f(t) \neq 0$ is true. Suppose for some pair of real numbers $a, b$ the condition $f(a) = f(b)$ is satisfied. By comparing the results obtained by substituting into $(\star)$ $(x, y) = (a, t)$ and $(x, y) = (b, t)$, we see that $2a^2f(t) = 2b^2f(t)$ must hold, from which we get $a = \pm b$.
Let $f(0) = k$. Substituting $(x, y) = (0, 0)$ into the equation $(\star)$, we see that
$f(k) = 0$ must be satisfied. Also, if we substitute $(x, y) = (0, k)$ into $(\star)$ and use
the fact $f(k) = 0$, we get $f(k^2) = 0$. Thus, we see that we must have $k^2 = \pm k$,
and therefore $k = 0, \pm 1$. Finally, substituting $(x, y) = (0, y)$ into $(\star)$, we get
$f(k - f(y)) = f(y^2)$, from which we can conclude that $k - f(y) = \pm y^2$ for any $y$,
i.e., for any $y$, $f(y) \in \{k + y^2, k - y^2\}$ must hold.

(1) Let us consider the case $k=0$.
Suppose there exist non-zero real numbers $z, w$ for which $f(z) = z^2$, $f(w) = -w^2$ are satisfied. By substituting $(x, y) = (w, w)$ into the equation $(\star)$, we obtain $0 = f(f(w)) + 2w^4 + f(w^2)$. Since $k = 0$, we have $f(f(w)) = \pm w^4$ and $f(w^2) = \pm w^4$, and we can check that $f(w^2) = -w^4$ must hold, since $w \neq 0$. Finally, if we substitute $(x, y) = (z, w)$ into the equation $(\star)$, we get
$$\pm(z^2 + w^2)^2 = \pm z^4 + 2z^2w^2 - w^4.$$
By simplifying each of the 4 cases arising from the combinations of ± signs,
we can check easily that there are no $(z, w)$'s with $z^2 > 0$, $w^2 > 0$ which
satisfies the last equation. We can thus conclude that if $k = 0$, then either
$f(x) = x^2$ for all $x$ or $f(x) = -x^2$ for all $x$.

(2) Consider the cases when $k = \pm 1$.
By substituting $(x, y) = (x, k)$ into the equation $(\star)$, we get
$$f(f(x) - k) = f(f(x)) - 2kx^2 + k.$$
From $f(x) - k = \pm x^2$, we have $f(f(x) - k) = k \pm x^4$, and since $f(f(x)) = k \pm f(x)^2$ we get
$$(\star\star) \qquad \pm f(x)^2 = \pm x^4 + 2kx^2 - k.$$
• Suppose $k = 1$ holds.
If there exists a $c \neq 0$ for which $f(c) = 1 + c^2$, then by substituting $x = c$ into the equation $(\star\star)$ above, we get
$$\pm(1 + c^2)^2 = \pm c^4 + 2c^2 = 1.$$
But we can check easily that this is impossible for any combination of ± signs, since $c^2 > 0$. Hence, if $k = 1$ we must have $f(x) = 1 - x^2$ for all $x$.

• Suppose $k = -1$ holds.
If there exists a $c \neq 0$ for which $f(c) = 1 - c^2$, we get from the equation $(\star\star)$ that
$$\pm(-1 - c^2)^2 = \pm c^4 - 2c^2 + 1.$$
But as for the case of $k = 1$, this identity (for any combination of ± signs) does not hold since $c^2 > 0$. Thus we must have $f(x) = -1 + x^2$ for all $x$.

Putting together the arguments made above, we conclude that there are only 5 possibilities: $0, x^2, -x^2, x^2 - 1, 1 - x^2$ for $f$ satisfying the equation $(\star)$.