$\underset{x\to 0}{lim} \frac{{\left(4^x-1\right)}^3}{\sin \left(\frac{x}{4}\right){\log }_e\left(1+\frac{x^2}{3}\right)}=$

$=\underset{x\to 0}{lim} \frac{{\left(4^x-1\right)}^3}{\sin \left(\frac{x}{4}\right){\log }_e\left(1+\frac{x^2}{3}\right)}$

Divide with x³ and simplify

$$\begin{gathered} = \frac{{\left(\underset{x\to 0}{lim} \frac{4^x-1}{x}\right)}^3}{\underset{x\to 0}{lim}\frac{\sin \left(\frac{x}{4}\right)}{{\displaystyle \frac{x}{4}}.4}\underset{x\to 0}{lim}{\log }_e\left(\frac{\left(1+\frac{x^2}{3}\right)}{{\displaystyle \frac{x^2}{3}}.3}\right)} \\ =12{\left({\log }_{e}4\right)}^3 \end{gathered}$$