Simplify:

$\left(3^{-1}+6^{-1}\right) \div\left(\frac{3}{4}\right)^{-1}$

$=\left(\frac{1}{3}+\frac{1}{6}\right) \div\left(\frac{4}{3}\right)^{1}$

$=\left(\left[\frac{1 \times 2}{3 \times 2}\right]+\left[\frac{1 \times 1}{6 \times 1}\right]\right) \div\left(\frac{4}{3}\right)$

$=\left(\frac{2+1}{6}\right) \div\left(\frac{4}{3}\right)$

$=\left(\frac{3}{6}\right) \div\left(\frac{4}{3}\right)$

$=\left(\frac{1}{2}\right) \div\left(\frac{4}{3}\right)$

$=\left(\frac{1}{2}\right) \times\left(\frac{3}{4}\right)$

$=\frac{3}{8}$

$\therefore\left(3^{-1}+6^{-1}\right) \div\left(\frac{3}{4}\right)^{-1}=\frac{3}{8}$