The equation itself: $$f(x) = f(3x) + \tanh(x)$$ So firstly I'm solving homogeneous equation: $$f(x)=f(3x)$$ so is just periodic function $\Theta(\ln x)$ with period $\ln 3$. So: $$F(x) = \Theta(\ln x) + \hat{f}(x)$$where $\hat{f}(x)$ is the particular solution of equation. Any tips how to find some?
Upd: New picture
It seems like its really bad around 0, but going smooth on big values. And my current guess that its behavior around $x=0$ is strongly connected to period of $\Theta$
For functions $f:(0,\infty)\to\mathbb{R}$, the general solution to the functional equation
$$f(x)=f(3x)+\tanh(x)$$
is
$$f(x)=\Theta(\ln x)-\sum_{k=1}^\infty\tanh\left(x\over3^k\right)$$
where $\Theta:\mathbb{R}\to\mathbb{R}$ is any function satisfying $\Theta(x+\ln3)=\Theta(x)$.
Note that the infinite series is absolutely convergent for any $x$, since $\tanh(x/3^k)\approx x/3^k$ for large enough $k$. It defines a continuous (indeed, smooth) function on all of $\mathbb{R}$. If you want the function $f(x)$ to have a limit as $x\to0^+$, then you need for $\Theta$ to be constant; otherwise $f$ will approach all the values $\Theta$ takes on (as in the OP's figure).