Solving functional equation $f(4x)-f(3x)=2x$

It's almost correct. You are correct that we can deduce that $$f(4x)=f(3x)+2x$$ and by repeatedly applying this we get $$f(4x)=f\left(4\left(\frac{3}4\right)^kx\right)+\sum_{n=0}^{k-1}2\left(\frac{3}4\right)^{n}x$$ You make an error on the next step, however. You mean to take a limit as $k$ goes to infinity, but you to this incorrectly. In particular, the correct expression would be: $$f(4x)=\lim_{k\rightarrow\infty}f\left(4\left(\frac{3}4\right)^kx\right)+8x$$ where we have a term of $\lim_{k\rightarrow\infty}f\left(4\left(\frac{3}4\right)^kx\right)$ that you omitted; in particular, this can be any constant, and the constant can be different for positive and negative numbers. It does, however, exist since $f$ is increasing. Thus, the solutions are of the form: $$f(x)=\begin{cases}2x+c_1&&\text{if }x>0\\c_2&&\text{if }x=0\\2x+c_3&&\text{if }x<0\end{cases}$$ for some constants $c_1\geq c_2 \geq c_3$.