Problem in solving functional equation $f(x^2 + yf(x)) = xf(x+y)$ To find all functions
$f$ which is a real function from $\Bbb R \to \Bbb R$ satisfying the relation 
$$f(x^2 + yf(x)) = xf(x+y)$$
 It can be easily seen that the identity function $i.e.$ $f(x)=x$ and $f(x)=0$ (verified just now) satisfies the above relation!! And putting $y=0$ I have got $f(x^2)=xf(x)$.
Help needed to find other functions satisfying the relation.
 A: 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 $x$.
A: 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 $x$.
A: Hint: Note that the relation is true for all $x,y$. Therefore
$$f(x^2) = xf(x)$$
this can be solved as:
$$f(x) = C x$$.
Note also that
$$f(x^2 -xf(x)) = xf(0) $$
A: An incomplete solution. We present two results on this problem.

We show that if $f(x)$ is injective then the only non-trivial solution to the functional equation is $f(x)=x$. 

Proof: Apart from the trivial solution $f(x)=0$ let us assume $f(x)$ is not identically zero and it is injective. Set $x=y=0$ to get 
$$f(0^2+0f(0))=0f(0)\Rightarrow f(0)=0$$
Then let $x=-y$ one would obtain
$$f(x^2-xf(x))=xf(x-x)=xf(0)\Rightarrow f(x^2-xf(x))=0$$
Using the assumption that $f(x)$ is injective then $$x^2-xf(x)=0\Rightarrow f(x)=x$$

$f(x)$ is an odd function.

Set $y=0$ then
$$f(x^2)=xf(x)$$
on the other hand 
$$f((-x)^2)=-xf(-x)\Rightarrow f(x^2)=-xf(-x)\Rightarrow f(x)=-f(-x)$$
and hence $f(x)$ would be an odd function. 
A: Putting $ x=0 $ gives $$f(yf(0))=0 $$ for any $y\in\mathbb{R}$ so $f(0)=0$. If $f$ is identically zero, then it satisfies the equation. So for non-constant solution we may assume $f$ is not zero at some point.  Setting $y=0$, $$ f(x^2)=xf(x). $$ Let $x_0$ be a real number for which $f(x_0)=0$. Then $$f(x_0 ^2)=x_0 f(x_0 +y)$$ and so $$ x_0 f(x_0 +y) =0 $$ for all $y\in\mathbb{R}$. Thus $x_0 =0$ since f is not identically zero. Indeed, 
$$f(x)=0 \iff x=0.$$
Now setting $ y=-x $ we have for any $x\in\mathbb{R}$,
$$f(x^2-xf(x))=0$$ and so 
$$x^2-xf(x)=0 $$ or 
$$f(x)=x.$$
