Answer
$\dfrac{\sqrt[3]{6xy^2}}{3x}$
Work Step by Step
Rationalizing the denominator of $
\dfrac{\sqrt[3]{2y^2}}{\sqrt[3]{9x^2}}
$ results to
\begin{array}{l}
\dfrac{\sqrt[3]{2y^2}}{\sqrt[3]{9x^2}}
\cdot
\dfrac{\sqrt[3]{3x}}{\sqrt[3]{3x}}
\\\\=
\dfrac{\sqrt[3]{6xy^2}}{\sqrt[3]{27x^3}}
\\\\=
\dfrac{\sqrt[3]{6xy^2}}{3x}
.\end{array}