Answer
$\frac{64}{27z^{3}}$
Work Step by Step
According to the definition of negative exponents, $a^{-n}=\frac{1}{a^{n}}$ (where $a\ne0$).
Therefore, $(\frac{3z}{4})^{-3}=\frac{1}{(\frac{3z}{4})^{3}}=(\frac{4}{3z})^{3}=\frac{4^{3}}{3^{3}z^{3}}=\frac{64}{27z^{3}}$.