$f: \Bbb{R} \to \Bbb{R}, f(-y^2)+f(x^2+y)-y=(x+y)f(x-y).$ 
$f: \Bbb{R} \to \Bbb{R}, f(-y^2)+f(x^2+y)-y=(x+y)f(x-y).$

From some Olympiad problem book.
My attempt:
\begin{align}
P(0, 0): \; & f(0)+f(0)=0, f(0)=0. \\
\ \\
P(x, x): \; & f(-x^2)+f(x^2+x)-x=0. \\
P(x, -x): \; & f(-x^2)+f(x^2-x)+x=0. \\
\therefore \; & f(x^2+x)-f(x^2-x)=2x. \\
\ \\
P(x, 0): \; & f(x^2)=xf(x). \\
\ \\
P(x, -x^2): \; & f(-x^4)+x^2=(x-x^2)f(x+x^2). \\
& =(x-x^2)(f(x^2-x)+2x) \\
& =-f(x^4-2x^3+x^2)+2x^2-2x^3.
\end{align}
I can't proceed more. Can anyone give some help or answer this question?

$$\begin{cases}P(x,y):\quad f(-y^2)+f(x^2+y)-y=(x+y)f(x-y)\\
P(-x,y):\quad f(-y^2)+f(x^2+y)-y=(-x+y)f(-x-y)
\end{cases}$$
Then $P(x,y)-P(-x,y)$ is $$(x+y)f(x-y)+(x-y)f(-x-y)=0.$$
Take $(a,b)\in\mathbb{R}^2$, and let $(x,y)$ be the solution of $$\begin{cases}x-y=a\\ x+y=-b \end{cases}$$
so that $P(x,y)-P(-x,y)$ becomes
$$-bf(a)+af(b)=0 \iff af(b)=bf(a).$$
You gave a formula for $P(x,-x)$, if you put $x=1$ in that formula you find $f(2)=2$.
Substituting $a=2$ in the formula above, $$2f(b)=bf(2)=2b,$$
meaning that $f$ is the identity.
