Problem in solving functional equation $f(x^2 + yf(x)) = xf(x+y)$

Solution 1:

Substitute $x=y=0$, we have $f(0)=0$.

Suppose $f(a)=0$ for some $a\ne 0$, then substitute $x=a$, gives $$f(a^2)=af(a+y) \hspace{1cm}\forall y$$ Hence we have the trivial solution $$f(y)=\text{constant}=f(a)=f(0)=0$$ for all $y$.

Therefore, if other solutions exist, they must satisfy $f(x)\ne0$ for all $x\ne0$.

Now for $x\ne 0$ let $$y=-x^2/f(x)$$ then $$f(0)=xf(x-x^2/(f(x)))=0$$

Hence we have $$x-x^2/f(x)=0$$ which implies $$f(x)=x$$ for all $x\ne 0$.

Combined with $f(0)=0$, the non-trivial solution is $$f(x)=x$$ for all $x$.

Solution 2:

Putting $x=y=0$: $$f(0)=0$$

Putting $y=-x$: $$ f(x^2-xf(x))=0$$

If there exists $a\neq 0$ such that $f(a)=0$, then put $x=a$: $$f(a^2)=af(a+y)$$

Then $f(x)$ is a constant, and easily find that $f(x)=0$.

If $f(x)=0$ iff $x=0$, then $x^2-xf(x)=0$ for all $x$, then $f(x)=x$ for all $x$.