$P(x,0): f(f(x))=xf(0)+f(x)$ $P(0,x): f(f(0))=f(0)$ $P(f(0),0): f(0)^2=0 \implies f(0)=0$ So, $f(f(x))=f(x)$. $P(f(x),-1): f(-1)=-1$ or $\boxed{f(x)=0}$. $P(x,-1): f(f(x)-x)=f(x)-x$ will finish later