Functional equation problem: $ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) $

Suppose we have $(x^2+1)^2+4f(x)\geq 0$ for some $x$. Then there exists some $y_0$, such that $$y_0^2-(x^2+1)y_0-f(x)=0.$$ We may also assume that $y_0\ne 0$, because the roots of the above quadratic can't both be $0$. Plugging $P(x, y_0)$ in the equation we get $$y_0f(x)^2=0,$$ and because $y_0\ne 0$, we must have $f(x)=0$.

This directly implies $f(x)\leq 0$ for all $x$. Suppose the function does not positive roots. This means that we must have $f(x)<\frac{-(x^2+1)^2}{4}$ for all positive $x$. If we plug $P(x, 0)$ in the equation, where $x$ is positive, we get $$f(0)=f(-f(x))<-\frac{(f(x)^2+1)^2}{4}<-\frac{\left(-\left(\frac{(x^2+1)^2}{4}\right)^2+1\right)^2}{4},$$ where the first inequality follows from $-f(x)>0$ and $f(x)<\frac{-(x^2+1)^2}{4}$, whereas the second one follows from the fact that squaring the inequality mentioned above implies $$f(x)^2>\left(\frac{-(x^2+1)^2}{4}\right)^2.$$ Seeing as the $RHS$ of the inequality is not bounded from below, we have reached a contradiction, therefore we must have a positive root $a$. Plugging $P\left(x, \frac{a}{x^2+1}\right)$ in the equation we get $$0\geq f\left(\left(\frac{a}{x^2+1}\right)^2-f(x)\right)=\frac{a}{x^2+1}f(x)^2\geq 0,$$ so $\boxed{f\equiv 0}$, which indeed is a solution.