Answer
$\dfrac{1}{3x}$
Work Step by Step
Simplify the radicand (expression inside the radicla sign) to obtain:
$$
\sqrt[3]{\frac{3xy^2}{81x^4y^2}}=\sqrt[3]{\frac{1}{27x^3}}==\sqrt[3]{\frac{1}{(3x)^3}}$$
Use the rule $\sqrt[n]{a^n}=a, a\gt0$ to obtain
$$\sqrt[3]{\frac{1}{(3x)^3}}=\frac{1}{3x}
$$
