Find all fucntions such that $f:\Bbb R \to \Bbb R$ and $f(x^{2}+yf(x))=f(x+y)$

Question

My work : first take $x=y=0$ we get $f(0)=f(0)=c$ , then take $x=x$ and $y=-x$ implies $f(x^{2}-xf(x))=f(0)$ now as there is another function inside the parenthesis I'm asuming there are some entries for which $x^{2}-xf(x)=0$ then we get $f(x)=x$. but I'm little confused about that assumption. If anyone can just fix my mistakes and solve that problem it'll be a great help for me.

Answer 1

Let $x=x$ and $y=0$ $ \implies f(x^2)=f(x)$

Let $x=-x$ $\implies f(-x)=f(x)$ i.e $f(x) $ is an even function

$f(x^2+yf(x))=f(x^2+yf(-x)$ since $f(x)=f(-x)$

$\implies f(x+y)=f(-x+y)$

If we let $y=-x$ then $f(0)=f(-2x)$ which implies that $f(x)=f(0)$. We are not given value of $f(0)$ nor we are able to find it. Since $f(0)$ is constant which means that $f(x)$ is an constant function.

Answer 2

As observed in the comments, $f(x^2)=f(x)$ for all $x$ so that $f(x)=f(-x)$ for all $x$. Thus, $f(x^2+yf(x))=f(x^2+yf(-x))$ for all $x$, and $f(x+y)=f(-x+y)$, for all $x,y$. This implies that $f$ is constant.