Let $g(x)=f(x)-1$. We see that the original statement becomes $P(x,y): y^2g(x)+x^2g(y)=xyg(x+y)$. Let $g(1)=a$. $P(x,1): g(x)=xg(x+1)-x^2a$ $P(x+1,1):g(x+1)=(x+1)g(x+2)-(x+1)^2a$ Eliminating $g(x+1)$ gives $g(x)+(x^3+3x^2+x)a=x(x+1)g(x+2)$---(1) $P(1,1): g(2)=2a$. $P(x,2): 4g(x)+x^2g(2)=2xg(x+2)$. $2(x+1)g(x)+(x^3+x^2)a=x(x+1)g(x+2)$ ---(2) Comparing (1) and (2) gives $g(x)+(x^3+3x^2+x)a=2(x+1)g(x)+(x^3+x^2)a$. $(2x+1)g(x) = (2x+1)xa$. For all $x \neq - \frac{1}{2}$, $g(x)=ax$. Then any random sub (such as $P( - \frac{1}{2} , - \frac{1}{2} )$)will give $g(x)=ax$ for $x=- \frac{1}{2}$ too. Thus, $f(x)=ax+1$.

let $f(x)=g(x)+!$then$y^2g(x)+x^2g(y)=xyg(x+y)$ let$y=0,g(0)=0$let$g(1)=c,h(x)=\frac{g(x)}{c}$then$h(1)=1,so h(2)=2$ then$h(x+1)=\frac{h(x)+x^2}{x};h(x+2)=\frac{h(x+1)+(x+1)^2}{x+1};h(x+2)=\frac{4h(x)+2x^2}{2x}$ solving this yields $h(x)=x$ hence$f(x)=cx+1$

Let $P(x,y)$ be the assertion into the problem statement. $P(1,0)\implies f(0)=1$. $P(1,-1)\implies f(-1)=2-f(1)$. $P(x,1)\implies$ \begin{align}f(x)+x^2f(1)+x=xf(x+1)+x^2+1\end{align}. $P(x+1,-1)\implies f(x+1)+(x+1)^2f(-1)-x-1=(-x-1)f(x)+(x+1)^2+1$ or \begin{align*} f(x+1)&=(-x-1)f(x)+x^2+3x+3-(x^2+2x+1)(2-f(1))\\&=(-x-1)f(x)-x^2-x+1+f(1)(x^2+2x+1).\end{align*}Substituting this in the into (1) yields $$f(x)(x^2+x+1)=-x^3+1+f(1)(x^3+x^2+x)$$so $f(x)=1+x(f(1)-1).$ But, $f(1)$ is not uniquely determined and can be any number so $\boxed{f(x)=ax+1}$ for some constant $a$. $\blacksquare$

$f(x)=g(x)+1$ Answer:$f(x)=ax+1$

Let $g(x) = f(x)-1$ so now we have $y^2g(x) + x^2g(y) = xyg(x+y)$. Let $g(1) = c$. $P(1,1) : 2g(1) = g(2) \implies g(2) = 2c$. By induction assume $g(n) = cn$. $P(n,1) : g(n) + n^2g(1) = ng(n+1) \implies nc + n^2c = ng(n+1) \implies g(n+1) = c(n+1)$ so we have $g(x) = cx$ which implies that $f(x) = cx + 1$.

Let $g(x)=f(x)-1$, then $P(x,y):x^2g(y)+y^2g(x)=xyg(x+y)$. $P(1,0)\Rightarrow g(0)=0$ $P(x,x)\Rightarrow 2x^2g(x)=x^2g(2x)\Rightarrow g(2x)=2g(x)$ since $g(0)=0$. In particular, $g(2)=2g(1)$. $P(x,-x)\Rightarrow x^2g(-x)+x^2g(x)=0\Rightarrow g$ is odd since $g(0)=0$. $P(x,1)\Rightarrow x^2g(1)+g(x)=xg(x+1)$ $P(x+1,1)\Rightarrow (x^3+3x^2+x)g(1)+g(x)=(x^2+x)g(x+2)$ $P(x,2)\Rightarrow x^2g(1)+2g(x)=xg(x+2)\Rightarrow (x^3+x^2)g(1)+(2x+2)g(x)=(x^2+x)g(x+2)=(x^3+3x^2+x)g(1)+g(x)$, so $(2x+1)g(x)=(2x^2+x)g(1)$ and $g(x)=xg(1)$ for all $x\ne-\frac12$. But since $g(1)=2f\left(\frac12\right)$ we have $\boxed{f(x)=cx-1}$.