Find $f'(0)$ if $f(x)+f(2x)=x\space\space\forall x$ Find $f'(0)$ if $f(x)+f(2x)=x\space\space\forall x$

If we assume $f'(0)\in\mathbb R$, then obviously, $f'(0)=\frac{1}{3}$.

But what if we don't assume the derivative exists?

I get this question when I am taking an exam, and I then asked the professor about whether there exist $f$ such that $f'(0)$ doesn't exist but satisfies the condition, but he says he think both exist or not is likely...

What I have tried:

If $f(x)+f(2x)=x\space\space\forall x\in\mathbb R$, $f(x)=\frac{x}{3}$ is the only nice function I have ever thought up$.

If we restrict $dom(f)\in\mathbb R$, I think that any polynomial besides $\frac{x}{3}$ doesn't satisfy the condition, and somethings like $\vert x\vert$ doesn't help neither.

On the other hand, I have tried to think about the equivalence statements with $f(x)+f(2x)=x\space\space\forall x$ in order to prove $f$ must be specific kind of problem s.t.$f'(0)$ exists.

$f(2x)+f(4x)=2x$, so $f(4x)-f(x)=x$.

In general, $$f(2^{2^n}x)-f(x)=x\prod_{k=1}^{n-1}(2^{2^k}+1)\forall n\in\mathbb N, x\in\mathbb R$$

But I don't think this helps.

Can we find $f(2x)-f(x)$ by given condition? I havn't get an idea. But while I am thinking of it, I observe that $g(x)=f(2x)-f(x)$ is on its own satisfying $g(2x)+g(x)=x$.

Will thinking about $f\circ f\circ f\circ f\circ f\circ\dots$ be useful? $f\biggl(f(x)+f(2x)\biggr)+f\biggl(2\bigl(f(x)+f(2x)\bigr)\biggr)=x\space\space\forall x$

Any help will be appreciate. Thank you!
Consider the function
$$
f(x)=\left\{\begin{array}{}\sin(\pi\log_2(|x|))+\frac x3&amp;\text{if }x\ne0\\ 0&amp;\text{if }x=0\end{array}\right.
$$
That is, the conditions given do not imply that $f'(0)$ exists.
$f(x)$ is separated into independent subsets defined by equivalence relationship $a = b$ given by $\exists k : 2^k \cdot a = b$.

This is due to two reorderings of the original equation, which can be iterated:

$$f(2x) = x - f(x)$$
$$f(x/2) = x/2 - f(x)$$

Let $f(1) = 1$ by definition. Then we have iteration:

0.5 -0.5
0.25 0.75
0.125 -0.625
0.0625 0.6875
0.03125 -0.65625
0.015625 0.671875
0.0078125 -0.6640625
0.00390625 0.66796875
0.001953125 -0.666015625


It is clear that there is no derivative at $f(0)$.