quacksaysduck wrote: Find all functions $f:\mathbb R\to\mathbb R$ such that \[f(x^2)+2xf(y)=yf(x)+xf(x+y).\] Let $P(x,y)$ be tghe assertion $f(x^2)+2xf(y)=yf(x)+xf(x+y)$ Let $c=f(1)$ $P(0,0)$ $\implies$ $f(0)=0$ $P(x,0)$ $\implies$ $f(x^2)=xf(x)$ (and so $f(-x)=-f(x)$) Adding $P(x,y)$ with $P(x,-y)$ and since $f(x^2)=xf(x)$, we get $2f(x)=f(x+y)+f(x-y)$ $\forall x\ne 0$, still true when $x=0$ This implies (using $f(0)=0$) that $f(x)$ is additive and then $P(x,y)$ becomes $xf(y)=yf(x)$ And so, setting there $y=1$ : $\boxed{f(x)=cx\quad\forall x}$, which indeed fits, whatever is $c\in\mathbb R$

Let $P(x,y)$ denote functional equation. $$P(0,0) \implies f(0)=0$$$$P(x, 0) \implies f(x^2)=xf(x)$$ $P(x, 1):$ $$xf(x)+2xf(1)=f(x)+xf(x+1)$$$P(1,x)$ $$f(1)+2f(x)=xf(1)+f(x+1)$$ After substituting expression for $f(x+1)$ in first equation and after some easy algebraic manipulation we get $$x(x+1)f(1)=(x+1)f(x)$$This means that for all $x \neq -1, f(x)=cx$. Beacuase of $P(-1, 0)$ this also works for $-1$ therefore $f(x)=cx$ is the solution.