$P(0,0)\Rightarrow f(0)=0$ $P(1,-1)\Rightarrow f(1)(f(-1)+1)=0$ If $f(1)=0$: $P(1,x-1)\Rightarrow\boxed{f(x)=0}$ If $f(-1)=-1$: $P(x,-1)\Rightarrow f(x^2)=xf(x)$, so $f$ is odd and $f(1)=1$. $P(x,y)\Rightarrow f(x+xy)=f(x)(1+f(y))$ since $f(0)=f(0)(1+f(y)$ $P(1,x)\Rightarrow f(x+1)=f(x)+1$ $P(x,y-1)\Rightarrow f(xy)=f(x)f(y)\Rightarrow f(x)^2=f(x^2)=xf(x)$ If $f(k)=0$ and $k\ne0$, then: $$f(x)=f\left(\frac xk\cdot k\right)=f\left(\frac xk\right)f(k)=0$$for all $x$, contradiction. So $\boxed{f(x)=x}$.

Answer is $\fbox{f(x)=0}$ and $\fbox{f(x)=x}$ Putting $x=0$ , we get $ \implies 0=f(0).f(y)$ If $f(y)=0$ , putting $y=x$ , we get $f(x)=0$ for all $x \in R$ which satisfy the main equation . Now let $f(x) \neq 0 $. So $f(0)=0$. Putting $x=1$ and $y=-1$ in the main equation we get $ \implies 0=f(1)[1+f(-1)]$. If $f(1)=0$ then putting $x=1$ in the main eqaution we get $f(1+y)=0$. Putting $1+y=x$ we get $f(x)=0$, a contradiction. So $f(-1)=-1$ Now putting $y=-1$ in the main equation we get $\implies f(0)=xf(x)+f(x^2)f(-1)$ $\implies f(x^2)=xf(x)$. Now $xf(x)=f(x^2)=-xf(-x)$ . So $f(-x)=-f(x)$. Thus $f(1)=1$ Putting $x=1$ in the main equation we get $\implies f(1+y)=1+f(y)$ Putting $y-1$ instead of $y$ we get $f(y)=1+f(y-1)$ Now putting $y-1$ instead of $y$ in the main equation we get $\implies xf(x+xy-x)=xf(x)+f(x^2)f(y-1)$ $\implies xf(xy)=xf(x)+xf(x)[f(y)-1]$ $\implies f(xy)=f(x)f(y)$ This is cauchy's multiplicative functional equation and solution to this equation is $f(x)=x^t$ where t is a constant. Now putting this in $f(x^2)=xf(x)$ we get $x^{2t}=x.x^t$ $\implies t=1$ So $f(x)=x$ is another solution which clearly satisfy the main equation