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Answer: The only such a function is $f(x) = 0$.
At first, assume that $f(x) > 0$ for some $x \in \mathbb{R}$. It means that we can choose $y$ such that
$$y^2 - f(x) = x^2y + y$$
(because for $f(x) > 0$ this equation has two solutions with respect to $y$), and if we insert it into the given equation we obtain an equality $yf(x)^2 = 0$. As $f(x) > 0$ then $y = 0$. But $y = 0$ is not a solution of $y^2 - f(x) = x^2y + y$ — contradiction. Thus $f(x) \le 0$ for all $x \in \mathbb{R}$.
Note that $f(x) = 0$ is a solution. So assume that $f(x_0) < 0$ for some $x_0 \in \mathbb{R}$. At first we show that $f$ is unbounded. Assume the contrary and put $x_0$ into the equation. We get that
$$f(y^2 - f(x_0)) - f(x_0^2y + y) = yf(x_0)^2,$$
and see that if $f(x)$ is bounded then the left hand side of this equality also is bounded, but the right hand side is unbounded, that is impossible.
If we put $y = 0$ in the original equation we get that $f(-f(x)) = f(0)$. As $f(x)$ is unbounded and nonpositive we conclude that we can find arbitrarily large $y$ such that $f(y) = f(0)$. Now put $x = x_0$ and choose $y_0$ such that $y_0 > \frac{-f(0)}{f(x_0)^2}$ and $f(y_0(x_0^2 + 1)) = f(0)$. We get that
$$f(y_0^2 - f(x_0)) = y_0f(x_0)^2 + f(y_0(x_0^2 + 1)) > -f(0) + f(0) = 0,$$
what contradicts the fact, that $f(x) \le 0$ for all real $x$.

For $y = 0$ we have $f(-f(x)) = f(0)$, in particular $f(-f(0)) = f(0)$. Denote $f(0) = c$.
For $x = 0$ we have $f(y^2 - c) = yc^2 + f(y)$ and substituting $-y$ instead of $y$ gives
$$f(y^2 - c) = -yc^2 + f(-y),$$
hence $f(-y) = 2yc^2 + f(y)$ for any $y$.
Finally,
$$yf(x)^2 + f(x^2y + y) = f(y^2 - f(x)) = f((-y)^2 - f(x)) = -yf(x)^2 + f(-x^2y - y) = \\ = -yf(x)^2 + 2(x^2y + y)c^2 + f(x^2y + y),$$
hence $2yf(x)^2 = 2(x^2y + y)c^2$ and $f(x) = \pm c\sqrt{x^2+1}$ (the choice of $\pm$ may depend on $x$).
Then $f(-f(0)) = f(0)$ gives $\pm c\sqrt{c^2+1} = c$, $c = 0$ and $f(x) = 0$ for every $x$.