ZETA_in_olympiad wrote: Find all functions $f:\mathbb R\to \mathbb R$ such that $$xy(f(x+y)-f(x)-f(y))=2f(xy)$$for all $x,y\in \mathbb R.$ Let $P(x,y)$ be the assertion $xy(f(x+y)-f(x)-f(y))=2f(xy)$ Let $c=f(1)$ $P(0,0)$ $\implies$ $f(0)=0$ $P(-2,1)$ $\implies$ $f(-1)=c$ $P(x-1,1)$ $\implies$ $f(x-1)=\frac{x-1}{x+1}f(x)-c\frac{x-1}{x+1}$ $\forall x\ne -1$ $P(x,-1)$ $\implies$ $f(x-1)=f(x)+c-\frac 2xf(-x)$ $\forall x\ne 0$ And so (subtracting) $\frac{x-1}{x+1}f(x)-c\frac{x-1}{x+1}=f(x)+c-\frac 2xf(-x)$ $\forall x\ne -1,0$ Which is $xf(x)-(x+1)f(-x)=-cx^2$ $\forall x\ne -1,0$, still true when $x=0$ or $x=-1$ Moving $x\to -x$, we get $-xf(-x)+(x-1)f(x)=-cx^2$ Cancelling $f(-x)$ between the two lines, we get : $\boxed{f(x)=cx^2\quad\forall x}$, which indeed fits, whatever is $c\in\mathbb R$

Clearly $f(0)=0.$ Let $P(x,y)$ denote the assertion. By $P(1,-1)$ we have $f(1)=f(-1).$ Comparing $P(x,1)$ and $P(x+1,-1)$ implies $$(\tfrac x{x+2})(f(x+1)-f(1))=f(x+1)+f(1)+\tfrac{2f(-x-1)}{-x-1} \quad (\dagger)$$Comparing $P(1,-x-2)$ and $P(-1,-x-1)$ with $(\dagger)$ gives $f(x+1)=(x+1)^2f(1);$ clearly works.