$P(x+1,x) \implies (2x+1)f(1)=f(x^2+2x+1-x^2) \implies f(2x+1)=(2x+1)f(1) \implies f(x)=cx$

Or $P(\frac {x+y} {2},\frac {x-y} {2}) \Rightarrow f(xy)=xf(y) \Rightarrow f(x)=xf(1) $

$P(x;-y)$ implies $f(x^2-y^2)=(x-y)f(x+y)=(x+y)f(x-y)$ so $f(x)=cx$

f(0)=0 so f(x)=xg(x) and change it. We get the amount, note that the g(x) must be constant.

We have $\frac{f({{x}^{2}}-{{y}^{2}})}{{{x}^{2}}-{{y}^{2}}}=\frac{f(x-y)}{x-y}$. Let $x+y=a, x-y=b$, so we get $\frac{f(ab)}{ab}=\frac{f(b)}{b}$, and now let $b=1$. So $f(a)=af(1)$.