Seeking a contest question on functional equation on $[-1,1]$ I vaguely remember a question going something like

Let $f$ be a function on $[-1,1]$ with $f$ satisfying (something like) 
  $$f(x^2-1)=(2x)f(x).$$
  Show that $f$ is identically zero on $[-1,1]$.

Sorry if I can't give much information. The exact statement has been bugging me for sometime now. I'd like to know what the exact statement is.
Edit: Swapped arguments.
Edit: Replaced $$f(x^2-1)=(2x-1)f(x).$$ with $$f(x^2-1)=2xf(x).$$.
 A: The left hand side of $f(x^2-1)=2x f(x)$ is an even function of $x$, so $f$ has to be odd. It follows that we can restrict to the interval $I:=[0,1]$ and have $f(1-x^2)=-2x f(x)$ there.
The map
$$T:\quad I\to I,\qquad x\mapsto Tx:= 1-x^2$$
is bijective and defines a discrete time dynamical system on $I$. Two points $x$, $y\in I$ belong to the same orbit iff $x=T^n y$ for some $n\in{\mathbb Z}$.
The point $\tau:={\sqrt{5}-1\over2}$ is a fixed point of $T$, and the set  $\{0,1\}$ is an orbit  of period $2$. Note that for $x\in\{0,\tau,1\}$ one necessarily has $f(x)=0$. Consider the map
$$S:=T^2:\quad x\mapsto 2x^2-x^4\ .$$
From $$Sx \ \cases{<x&$(0< x<\tau)$\cr =x &$(x\in\{0,\tau,1\})$\cr >x&$(\tau<x<1)$\cr}$$
we conclude that $0$ and $1$ are attracting fixed points of $S$ whereas $\tau$ is repelling. This implies that the sets $\{\tau\}$ and $\{0,1\}$ are the only finite orbits of $T$. Put $f(x):=0$ for $x\in\{0,\tau,1\}$. Then choose a point $x_\alpha$ in each infinite orbit $O_\alpha$, put $f(x_\alpha):=1$ (or some arbitrary value), and use the functional equation $f(Tx)=-2x f(x)$ to define $f$ on all of $O_\alpha$. The resulting $f:I\to{\mathbb R}$ will be $\ne0$ at most points of $I$.
This construction shows that we need additional assumptions on $f$ to guarantee $f(x)\equiv0$. The following heuristic argument makes plausible that $f(x)\equiv0$, if we assume that $f$ is differentiable at $0$.
The functional equation $f(Tx)=-2x f(x)$ implies
$$f(Sx)=f(2x^2-x^4)=4x(1-x^2) f(x)\ .$$
For $x$ near $0$ this "can be replaced" by $f(2x^2)= 4x f(x)$. We now consider the function
$$g(x):={f(x/2)\over x/2}={1\over2}{f(x^2/2)\over x^2/2}={1\over 2}g(x^2)\ .$$
It follows that for all $n\geq1$ we have
$$g(x)={1\over2^n}g\bigl(x^{2^n}\bigr)\ .$$
As $n\to\infty$ the right side converges to $0\cdot f'(0)=0$, from which we dare to conclude that $g(x)=0$ for all $x$ sufficiently near $0$, whence $f(x)=0$ for these $x$. Using the fact that the iterates of $S$ push all $x<\tau$ towards $0$ it follows that in fact $f(x)=0$ for $0\leq x<\tau$, and applying $T$ once gives the claim for the interval $]\tau,1[\ $.
A: First, you have: $f((\frac{1}{2})^2-1)=f(-\frac{3}{4})=(2\times\frac{1}{2}-1)f(\frac{1}{2})=0$. Then, let's consider $\varphi(x)=x^2-1$ and $(x_n)$ defined by $u_0=-\frac{3}{4}$ and $u_{n+1}=\varphi(u_n)$. It can easily be shown that $\forall n\in\mathbb{N}, f(u_n)=0$ (by recursion). 
It can also be shown that $\liminf u_n=-1$ and $\limsup u_n=0$, which means that  if $f$ is continuous, then by continuity $f(0)=0$ and $f(-1)=0$.
If $f$ is also monotonic, then $f(x)=0\space\forall x\in[-1,0]$. Now, we have $f(1^2-1)=1\times f(1)$, so $f(1)=0$ as well, so because of the monotony, $f(x)=0 \text{   }\forall x\in [-1,1]$.
Maybe someone can find weaker conditions as well...
