Answer
$\frac{1}{16}$
Work Step by Step
We know that $(a)^{-\frac{m}{n}}=\frac{1}{(a)^{\frac{m}{n}}}$ (as long as $(a)^{\frac{m}{n}}$ is a real number).
Therefore, $64^{-\frac{2}{3}}=\frac{1}{64^{\frac{2}{3}}}$
We also know that $(a)^{\frac{m}{n}}=(\sqrt[n] a)^{m}$ (where m and n are positive integers greater than 1 with $\frac{m}{n}$ in simplest form and $\sqrt[n] a$ is a real number).
$\frac{1}{64^{\frac{2}{3}}}=\frac{1}{(\sqrt[3] 64)^{2}}=\frac{1}{4^{2}}=\frac{1}{4\times4}=\frac{1}{16}$
We know that $\sqrt[3] 64=4$, because $4^{3}=64$
