Let $P(x,y)$ be the given assertion, $P(0,0)\implies f(0)=0$. Evaluating $P(x,x),P(2x,x),P(3x,x),\ldots$ will gives us $f(nx)=n^2f(x)$ by induction where $n\in \mathbb{N}$ and in particular, $f(n)=n^2f(1)$. Let $x=\frac{1}{n}$, we then get $f\left(\frac{1}{n}\right)=f(1)\frac{1}{n^2}$. So, $f\left(\frac{m}{n}\right)=m^2f\left(\frac{1}{n}\right)=f(1)\left(\frac{m}{n}\right)^2\ \ \forall m,n\in \mathbb{N}$. Also, \[P(x,-2x) \implies f(-x)=-8f(x)+f(3x)=f(x)\]so $f$ is an even function. Now, let $x,y\in\mathbb{Q}_{\geq 0}$ and WLOG $f(1)\geq 0$, \[P(x,x+y)\implies f(2x+y)=4yf(x)+f(y)=4f(1)yx^2+f(y)\geq f(y).\]So, $f$ is a montone function which implies continuity, thus $f(x)=f(1)x^2$ and it is easy to see that this satisfies the equation. $\blacksquare$

Let $P(x,y)$ be the given assertion. $P(0,1)$ gives $f(0)=0,$ then $P(x,x)$ gives $f(2x)=4f(x)$. Now $P(2x,x)$ gives $f(3x)=9f(x),$ and so $P(x,2x)$ gives $f(-x)=f(x).$ Finally, the expression $P(x,1)-xP(1,x)$ gives $f(x)=x^{2}f(1)\;\forall \;x\in \mathbb{R},$ which is indeed a solution for every choice of $f(1)\in \mathbb{R}.$