Solve functional equation find all  functions $f:\mathbb{R^{*}}\to \mathbb{R}$ such that
 $$f(y^2f(x)+x^2+y)=x(f(y))^2+f(y)+x^2,\;\forall x,y\in \mathbb{R^{*}}$$
($\mathbb{R^{*}}=\{x\in\mathbb{R},x\ne 0\})$
 A: 
does anyone find the mistake? (there must be one) remark: i'm assuming $f$ to be continuously extendable in $0$

I'm assuming $f$ to be once continously differentiable on it's connected domain in $\mathbb{R}$ and $0$ is contained in this domain.
$
\forall x,y : \frac{f\left(y^{2}f\left(x\right)+x^{2}+y\right)-f\left(y\right)}{x}=f\left(y\right)^{2}+x
$
$
\underbrace{\Rightarrow}_{y\rightarrow 0} \forall x: \frac{f\left(x^{2}\right)-f\left(0\right)}{x}=f\left(0\right)^{2}+x
$
this gives
$
\underbrace{\Rightarrow}_{x\rightarrow 0} 2f'\left(0\right)=f\left(0\right)^{2}
$
and
$
\forall x:f\left(x^{2}\right)-f\left(0\right)=xf\left(0\right)^{2}+x^{2}
$
$
\underbrace{\Rightarrow}_{\partial_{x}} \forall x: 2f'\left(x\right)=f\left(0\right)^{2}+2x
$
from there we see firstly, $f$ is parabolic (and also twice differentiable)
secondly
$
\underbrace{\Rightarrow}_{x\rightarrow 0} f\left(0\right)=0 \underbrace{\Rightarrow}_{2f'\left(0\right)=f\left(0\right)^{2}} f'\left(0\right)=0
$
and thirdly by applying $\partial_{x}$ again
$
2f''\left(x\right)=2
$
thus $f$ is parabolic and we have $f'\left(0\right)=0$, $f\left(0\right)=0$ and $f''\left(0\right)=1$.
this means $f\left(t\right)=\frac{1}{2}t^{2}$
A: The correct solution to the functional equation that doesn't exclude $x=0$ from its domain, that is $$f(y^2f(x)+x^2+y)=x(f(y))^2+f(y)+x^2,\;\forall x,y\in \mathbb{R},$$
can be found as follows.
Letting $y=0$, the functional equation reduces to:
$$f(x^2)=x(f(0))^2+f(0)+x^2$$
Set $f_0=f(0)$, and substitute $x=\text{sgn}{(u)}\sqrt{u}$. Then $u=x^2$, and
$$f(u) = u + f_0 + f_0^2 \text{sgn}{(u)}\sqrt{u}$$.
Letting $u=0$, 
$$f_0=f_0+f_0^2\\
\implies f_0^2=0\\
\implies f_0=0.$$
Hence, the general solution is simply the identity map, $f(x)=x$. It's a simple matter to check that this satisfies the functional relation. On the LHS we have,
$$f(y^2f(x)+x^2+y)=f(y^2x+x^2+y)\\
=y^2x+x^2+y.$$
And on the RHS we have,
$$x(f(y))^2+f(y)+x^2=xy^2+y+x^2.$$
