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.
 A: Let $f:\mathbb{R}\to\mathbb{R}$ and for $x,y\in\mathbb{R}$ denote by $P(x,y)$ the assertion $f(xf(x)+f(y))=x^2+y$. Assume that $f$ satisfies $P(x,y)$ for all $x,y\in\mathbb{R}$. Then
$$
P(0,y):\quad f(f(y))=y\implies f\text{ bijective}\\
P(f(x),y):\quad f(f(x)f(f(x))+f(y))=f(x)^2+y\implies x^2+y=f(xf(x)+f(y))=f(f(x)f(f(x))+f(y))=f(x)^2+y\\
\implies f(x)^2=x^2
$$
We are no left to prove that either $f(x)=x$ for all $x$, or $f(x)=-x$ for all $x$, i.e. that $f$ doesn't jump around between $x\mapsto x$ and $x\mapsto -x$. For this, assume that there are $a,b\in\mathbb{R}\backslash\{0\}$ with $f(a)=a$ and $f(b)=-b$. Then
$$
P(a,b):\quad f(a^2-b)=a^2+b
$$
Now if $f(a^2-b)=a^2-b$ then $a^2-b=a^2+b$ and thus $b=0$, contradiction. But if $f(a^2-b)=-(a^2-b)$ then $-(a^2-b)=a^2+b$ and thus $a=0$, so again contradiction. Therefore, such $a,b$ don't exist, and thus either $f(x)=x\ \forall x$ or $f(x)=-x\ \forall x$, and one can verify easily that this are indeed solutions to the equation.
A: Plugging in $x=y=-1$ shows that $f(0)=0$ and plugging in $x=0$ then shows that $f(f(y))=y$. This implies that $f$ is invertible, and plugging in $y=0$  shows that
$$f(xf(x))=x^2=f(f(x^2)),$$
and hence $xf(x)=f(x^2)$ for all $x\in\Bbb{R}$. This shows that $f(-x)=-f(x)$ for all $x\in\Bbb{R}$, and that
$$f(x^2+y)=xf(x)+f(y)=f(x^2)+f(y),\tag{1}$$
for all $x,y\in\Bbb{R}$, from which it follows that $f$ satisfies Cauchy's functional equation
$$f(x+y)=f(x)+f(y).$$
Much has been said about this functional equation, which has many pathological solutions. Note that this means $f$ is $\Bbb{Q}$-linear, and if $f$ is either continuous at a point, bounded on an interval or monotonic on an interval, then $f$ is $\Bbb{R}$-linear and so $f(x)=cx$ for some $c\in\Bbb{R}$.
In an earlier version of this answer I rushed to the conclusion that $f(x)=cx$, which quickly implies that $c=\pm1$ and indeed both functions $f(x)=\pm x$ satisfy the functional equation.
A: $$ \begin{align}
&\text{As already noted, we have }\qquad\qquad\qquad\qquad\qquad f(f(y)) = y \implies f(x) = f^{-1}(x) \\
&\text {Also, using the substitution } x\to y \text{ we get}\,\,\quad f((y+1)f(y)) = y^2+y \to f(0) = 0\\
&\text{using } f= f^{-1}:\quad f(x*f(x) + f(y)) = x^2+y \implies x*f(x) + f(y) = f(x^2+y)\\
&\text{from above line we then get}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad f(x) = \frac{f(x^2+y) -f(y)}{x}
\end{align}$$
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 $f(x^2) = f(f(x)^2) \implies x^2 = f(x)^2 \implies f(x) = \pm x$
