Find all functions $f:\Bbb{R} \to \Bbb{R}$ such that for all $x,y,z \in \Bbb{R} $ , $f(xf(x)+f(y))=x^2+y$ 
Find all functions $f:\Bbb{R} \to \Bbb{R}$ such that for all $x,y, \in \Bbb{R} $ , $f(xf(x)+f(y))=x^2+y$ 

We can easily get a strong condition $f(f(y))=y $ by setting  $x=0$ . By this equation  we know $f$ is injective and surjective.  I got lost from there. By  observation  I know $f(x)=x $ and $f(x)=-x$  are solution. So I was trying to make $x^2+y=f(xf(x)+f(y))$ close to $f(x)^2+y$ or $x^2+f(y)$. Any hints would be helpful.

$$ \begin{align}
&\text{As already noted, we have }\qquad\qquad\qquad\qquad\qquad f(f(y)) = y \implies f(x) = f^{-1}(x) \\
&\text {Also, using the substitution } x\to y \text{ we get}\,\,\quad f((y+1)f(y)) = y^2+y \to f(0) = 0\\
&\text{using } f= f^{-1}:\quad f(x*f(x) + f(y)) = x^2+y \implies x*f(x) + f(y) = f(x^2+y)\\
&\text{from above line we then get}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad f(x) = \frac{f(x^2+y) -f(y)}{x}
\end{align}$$
From $f(x) = \frac{f(x^2+y) -f(y)}{x}$ we now obtain $xf(x) = f(x^2)$ by setting $y=0$.
A substitution $x\mapsto f(x)$ now yields $f(x)x = f(f(x)^2)$.
Combining those two equations, we get $f(x^2) = f(f(x)^2) \implies x^2 = f(x)^2 \implies f(x) = \pm x$
