Prove $f(x)=0$ when $f(2x^2-1)= 2xf(x)$

Put $x=\cos(u)$, we get $f(\cos(2u))=2\cos(u)f(\cos(u))$, and by induction that $$f(\cos(2^nu))=2^n\prod_{k=0}^{n-1}\cos(2^ku))f(\cos(u))$$ We multiply by $\sin(u)$, we use $2\sin(u)\cos(u)=\sin(2u)$, and we get that $\sin(u)f(\cos(2^nu))=\sin(2^nu)f(\cos(u))$.

Now let $k$ an integer, choose $\displaystyle u=(\frac{2k+1}{2^n})\pi/2$. Then as $f(0)=0$, we get $\displaystyle f(\cos((\frac{2k+1}{2^n}))\pi/2)=0$.Now if $A=\{\frac{2k+1}{2^n}, k\geq 0, n\geq 0\}$, we know that $A\cap [0,2]$ is dense in $[0,2]$. Hence by continuity we get that $f(\cos(x\pi/2))=0$ for $x\in [0,2]$, and we are done.

Let $g : \Bbb{R} \to \Bbb{R}$ be defined by $g(\theta) = f(\cos\theta)$. Then it suffices to prove that $g \equiv 0$ on $[0, \pi]$.

From the functional equation for $f$, we find that

$$g(2\theta) = 2\cos(\theta)g(\theta) \tag{1}.$$

Now we consider the set $\mathcal{D}$ defined by

$$ \mathcal{D} = \bigg\{ \frac{2\pi k}{2^n + 1} : \text{$n \geq 1$ and $k = 1, \cdots, 2^{n-1}$ are integers} \bigg\}. $$

Notice that $\mathcal{D}$ is dense in $[0, \pi]$. Now for each $\theta = 2\pi k/(2^n + 1) \in \mathcal{D}$, we have $\sin \theta \neq 0$ and

$$ g(2^n\theta) = 2^n \cos(2^{n-1}\theta) \cdots \cos(\theta) g(\theta) = \frac{\sin(2^n \theta)}{\sin(\theta)}g(\theta). \tag{2} $$

Since $2^n \theta = 2\pi k - \theta$, we have $\cos(2^n\theta) = \cos(\theta)$ and $\sin(2^n\theta) = -\sin(\theta)$. Plugging this to $\text{(2)}$ shows that $g(\theta) = 0$ for $\theta \in \mathcal{D}$. Then by the density of $\mathcal{D}$ and the continuity of $g$, we have $g\equiv 0$ as desired.