any ideas?

Let $P(x,y): f (f(x) + f(y)) = f(x^2) + 2x^2 f(y) + (f(y))^2$. $P(x,0): f(f(x)+f(0))=f(x^2)+2x^2f(0)+(f(0))^2$. $P(0,x): f(f(x)+f(0))=f(0)+(f(x))^2$. Thus $f(x^2)+2x^2f(0)+(f(0))^2=f(0)+(f(x))^2$ So $P(x,y)-P(y,x): x^2f(y)-x^2f(0)=y^2f(x)-y^2f(0)\Leftrightarrow \dfrac{f(y)-f(0)}{y^2}=\dfrac{f(x)-f(0)}{x^2}=C$ therefore $f(x)=Cx^2+D$ $.....\Rightarrow f(x)=x^2,\forall x\in \mathbb{R},f(x)\equiv 0$ the solutions.

Prod55 wrote: Let $P(x,y): f (f(x) + f(y)) = f(x^2) + 2x^2 f(y) + (f(y))^2$. $P(x,0): f(f(x)+f(0))=f(x^2)+2x^2f(0)+(f(0))^2$. $P(0,x): f(f(x)+f(0))=f(0)+(f(x))^2$. Thus $f(x^2)+2x^2f(0)+(f(0))^2=f(0)+(f(x))^2$ So $P(x,y)-P(y,x): x^2f(y)-x^2f(0)=y^2f(x)-y^2f(0)\Leftrightarrow \dfrac{f(y)-f(0)}{y^2}=\dfrac{f(x)-f(0)}{x^2}=C$ therefore $f(x)=Cx^2+D$ $.....\Rightarrow f(x)=x^2,\forall x\in \mathbb{R},f(x)\equiv 0$ the solutions. Nice solution