Find all functions such that $f(x^2+y)=xf(x)+f(y)$ 
Find all functions (over real numbers) such that
  $$f(x^2+y)=xf(x)+f(y).$$

My idea: Put $x=1$. Therefore the given equation becomes $f(y+1)=f(y)+f(1)$. It is in the Cauchy's first form, so we get $f(x)=cx$. It also satisfies that given functional equation.
Is my answer correct?? If not then what is the complete solution?? Please help me!
 A: Here is a proof that $f$ does satisfy Cauchy's functional equation.  For now, I do not have a complete answer, unless some sort of continuity conditions is given.
Let $P(x,y)$ denote the condition that $f\left(x^2+y\right)=x\,f(x)+f(y)$.  Then, $P(1,0)$ implies that $f(0)=0$.  Now, $P(x,0)$ implies that $$f\left(x^2\right)=x\,f(x)$$ for all $x\in\mathbb{R}$.  This shows that $$f\left(x^2+y\right)=f\left(x^2\right)+f(y)$$ for all $x,y\in\mathbb{R}$.  Next, $P\left(x,-x^2\right)$ leads to $f\left(-x^2\right)=-f\left(x^2\right)$, and consequently, $$f(-x)=-f(x)$$ for every $x\in\mathbb{R}$.  From this, it can be easily shown that $f$ satisfies Cauchy's functional equation.
Per Mohsen shahriari's comment below, knowing that $f$ satisfies Cauchy's functional equation implies that $f(x)=f(1)\,x$ for all $x\in\mathbb{R}$.  Here is a paraphrase of the comment in case it is deleted.
Let $x\in\mathbb{R}$.  Note that
$$(x+1)\,f(x+1)=(x+1)\,\big(f(x)+f(1)\big)=(x+1)\,f(x)+(x+1)\,f(1)\,,\tag{1}$$
that
$$\begin{align}(x+1)\,f(x+1)&=f\big((x+1)^2\big)=f\big(x^2+(2x+1)\big)\\&=x\,f(x)+f(2x+1)=x\,f(x)+2\,f(x)+f(1)\\&=(x+2)\,f(x)+f(1)\,.\tag{2}\end{align}$$
From (1) and (2),
$$(x+1)\,f(x)+(x+1)\,f(1)=(x+2)\,f(x)+f(1)\,,$$
whence
$$f(x)=x\,f(1)\,.$$
A: Assuming $f \in C$ and arranging as
$$
\frac{f(y+x^2)-f(y)}{x^2}=\frac{f(x)}{x} = \phi(x)
$$
So
$$
\lim_{x\to 0}\frac{f(y+x^2)-f(y)}{x^2}=\phi(0)
$$
which is independent of $y$ hence $f(x) = C_0 x$
