Find all functions $f$ such that $f:\mathbb{R}\rightarrow \mathbb{R}$ and $f\big(f(x)-x+y^2\big)=yf(y)$. 
Find all functions $f$ such that $f:\mathbb{R}\rightarrow \mathbb{R}$ and $$f\big(f(x)-x+y^2\big)=yf(y)\,.$$



*

*If $y=0$ we get $f(f(x)-x)=0$.

*If $b=f(1)$, then putting $x=y=1$ we get $\boxed{f(b)=b}$, so $f$ has a fixed point.

*If $x=b$ we get $\boxed{f(y^2)=yf(y)}$ for all $y$, so pluging $y=0$ we get $\boxed{f(0)=0}$.

Nothing usefull really. Any idea how to do it?
 A: Suppose $f:\Bbb R\to \Bbb R$ satisfies
$$f\big(f(x)-x+y^2\big)=yf(y)\tag{1}$$
for all $x,y\in \Bbb R$.  Plugging in $x,y=1$ in $(1)$, we get
$$f\big(f(1)\big)=f(1).$$
Hence, substituting $x=f(1)$ in $(1)$, we obtain
$$f(y^2)=yf(y).\tag{2}$$
This shows that $f(0)=0$ and for $y\ne 0$,
$$-yf(-y)=f\big((-y)^2\big)=f(y^2)=yf(y),$$
so $f(-y)=-f(y)$.  Hence $f$ is an odd function.
From $(1)$ and $(2)$, we get
$$f\big(f(x)-x+y^2\big)=f(y^2)$$
for all $x,y\in\mathbb{R}$. Substitute $-x$ for $x$ in the last equation and using the fact that $f$ is odd, we get
$$f\big(f(x)-x-y^2\big)=f(-y^2)$$
for all $x,y\in\mathbb{R}$.  Hence,
$$f\big(f(x)-x+y)=f(y)\tag{3}$$
for all $x,y\in\mathbb{R}$.
Define $P$ to be the additive subgroup of $\mathbb{R}$ generated by $\big\{f(x)-x\big|x\in\Bbb R\big\}$.  Then, we see that for any $z\in\mathbb{R}$ and $p\in P$, we have
$$f(z+p)=f(z).\tag{4}$$
Consequently $f$ is an odd periodic function which is invariant under translation by elements of $P$. 
We claim that $P=f^{-1}(0)$.  First if $p\in P$, then $f(p)=f(p+0)=f(0)=0$, so $p\in f^{-1}(0)$.  Conversely, suppose that $p\in f^{-1}(0)$, then with $x=p$ in $(3)$, we have
$$f(-p+y)=f\big(f(p)-p+y\big)=f(y)$$
so that $p\in P$. 
For each $p\in P$, we see that
$$f\left(\left(\frac{1+p}{2}\right)^2\right)=f\left(\left(\frac{1-p}{2}\right)^2+p\right)=f\left(\left(\frac{1-p}{2}\right)^2\right)$$
From $(2)$, we have
$$\frac{1+p}{2}f\left(\frac{1+p}{2}\right)=\frac{1-p}{2}f\left(\frac{1-p}{2}\right).$$
However
$$f\left(\frac{1+p}{2}\right)=f\left(\frac{1-p}{2}+p\right)=f\left(\frac{1-p}{2}\right).$$
This means
$$f\left(\frac{1+p}{2}\right)=0$$
for all $p\in P$ such that $p\ne 0$.      Hence, for $p\in P\setminus\{0\}$, $\frac{1+p}{2}\in P$, so $1+p \in P$, making $1\in P$.  This shows that either $P=\{0\}$ or $\mathbb{Z}\left[\frac12\right]\subseteq P$.
If $P=\{0\}$, then $f(x)-x=0$ for all $x\in \Bbb R$.  Therefore, $f(x)=x$ for every $x\in \Bbb R$.  This yields one solution of $(1)$.  From now on $P\neq \{0\}$.  Thus $\mathbb{Z}\left[\frac12\right]\subseteq P$.  We want to show that $P=\Bbb R$, so that $f(x)=0$ for every $x\in \Bbb R$, and this is another solution of $(1)$.
For an arbitrary $y\in\mathbb{R}$, we see that
$$f\left(\left(y+\frac{1}{2}\right)^2\right)=\left(y+\frac{1}{2}\right)f\left(y+\frac{1}{2}\right)=\left(y+\frac{1}{2}\right)f(y),$$
since $1/2\in P$.  That is
$$f\left(\left(y+\frac12\right)^2\right)=\frac{1}{2}f(y)+yf(y)=\frac12f(y)+f(y^2).$$
Consequently
\begin{align}f\left(\left(y+\frac12\right)^2\right)-\left(y+\frac12\right)^2&=\frac{1}{2}f(y)+f(y^2)-\left(y+\frac12\right)^2\\&=\left(\frac{1}{2}f(y)-y\right)+\left(f(y^2)-y^2\right)+\frac14.\end{align}
Because $f\left(\left(y+\frac12\right)^2\right)-\left(y+\frac12\right)^2$, $f(y^2)-y^2$, and $\frac14$ are all elements of $P$, we conclude that
$$\frac{1}{2}f(y)-y\in P.$$
Therefore,
$$\big(f(y)-y\big)-y=2\left(\frac{1}{2}f(y)-y\right)\in P.$$
As $f(y)-y\in P$, we get $y\in P$.
