Let $P(x,y)$ denote assertion of given functional equation. Note that $P(x,0)$ gives us $f(-f(x)) = f(0)$. Comparing $P(x,y)$ and $P(x,-y)$ yields: $$yf(x)^2+f(x^2y+y) = -yf(x)^2 + f(-x^2y - y) \implies 2yf(x)^2 = f(-x^2y-y) - f(x^2y+y) \qquad (1) $$In $(1)$ setting $x = 0$ yields: $$ 2yf(0)^2 = f(-y)-f(y)$$Therefore: $$2yf(x)^2 = f(-x^2y-y) - f(x^2y+y) = 2(x^2y+y)f(0)^2 \implies f(x)^2 = (x^2+1)f(0)^2 \qquad (2)$$If $f(0) =0$ from $(2)$ we get that: $$f(x)^2 =0 \implies f(x) =0$$Otherwise ( if $f(0) \ne 0 $) in (2) replacing $x$ by $-f(x)$ gives us: $$f(-f(x))^2 =( f(x)^2+1)f(0)^2 = f(0)^2 \implies f(x)^2 + 1 = 1 \implies f(x)=0$$for all real numbers $x$, which is clearly impossible since $f(0) \ne 0$ We conclude that $f(x) =0$ for all real numbers $x$