Consider,

$\underset{x\to -\infty }{\lim }\frac{3\left|x\right|-x}{\left|x\right|-2x}=\underset{\mathrm{x}\to -\infty }{\lim }\frac{3\left(-x\right)-x}{\left(-x\right)-2x}\left[\therefore x\to -\infty ,\left|x\right|=-x\right]$

$=\underset{x\to -\infty }{\lim }\frac{-3x-x}{-x-2x}$

$=\underset{x\to -\infty }{\lim }\frac{-4x}{-3x}$

$=\frac{4}{3}$

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

$=\frac{4-3}{3}$

$=\frac{1}{3}$