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 $Sx \ \cases{x&$(\tau 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[\ $.