$f(x^2+yf(x))=xf(x+y)$.Find f

Question

Find all real functions $f:R\mapsto R$ satisfying the relation $$f(x^2+yf(x))=xf(x+y).$$

My Answer: Putting $y=0$ we get, $f(x^2)=xf(x)$. o

Which implies $\frac{f(x^2)}{x^2}=\frac{f(x)}{x}$.

Let $g(x)=\frac{f(x)}{x}$.(Assume $x\ne 0$)

Hence, $g(x)=g(x^2)$.

Therefore, $g(x)=c$, Or $f(x)=cx$ (when $x \ne 0$)

And $f(0)=0$ (By putting $x=y=0$).

Is my answer is ask right?? If not then kindly tell me the mistake!

Answer 1

Hint: if $f$ is not identically 0 then prove that $f(x)=0$ if and only if $x$ is 0. Then show that $f(x^2-xf(x))=xf(0)=0$. Thus $f(x)=x$.