Given $\underset{x\to 0}{\lim }\frac{{\sin }^{-1}x-{\tan }^{-1}x}{3x^3}$

$\Rightarrow \underset{x\to 0}{\lim }\frac{\left(x+\frac{x^3}{3!}\dots \right)-\left(x-\frac{x^3}{3}\dots \right)}{3x^3}$

$\Rightarrow \underset{x\to 0}{\lim }\frac{\left(1+\frac{3x^2}{3!}\dots \right)-\left(1-\frac{3x^2}{3}\dots \right)}{9x^2}$ (Use L'Hospital rule $\frac{0}{0}$ form)

$\Rightarrow \underset{x\to 0}{\lim }\frac{\left(\frac{6x}{3!}+\dots \right)-\left(-\frac{6x}{3}+\dots \right)}{18x}$  (Use L'Hospital rule $\frac{0}{0}$ form)

$\Rightarrow \underset{x\to 0}{\lim }\frac{\left(\frac{6}{3!}+\dots \right)-\left(-\frac{6}{3}+\dots \right)}{18}=\frac{1+2}{18}=\frac{1}{6}$

Given that, $\underset{x\to 0}{\lim }\frac{{\sin }^{-1}x-{\tan }^{-1}x}{3x^3}=L$

$\therefore \frac{1}{6}=L\Rightarrow 6L=1$

So, $6L+1=2$