Answer
$3^{-1}x^{-1}=\frac{1}{3x}$
Work Step by Step
We use the rules of exponents and radicals to obtain:
$\displaystyle \sqrt[3]{\frac{3xy^2}{81x^4y^2}}=\sqrt[3]{\frac{1}{27}x^{1-4}y^{2-2}}=\sqrt[3]{27^{-1}x^{-3}y^0}=\sqrt[3]{(3^{-3})x^{-3}}=3^{-1}x^{-1}=\frac{1}{3x}$
