Answer
$\displaystyle \frac{2x}{\sqrt[3]{2x^{2}y}}$
Work Step by Step
$ \displaystyle \frac{\sqrt[3]{4x}}{\sqrt[3]{}y}= \frac{\sqrt[3]{2^{2}x}}{\sqrt[3]{}y}$
... We want $\sqrt[3]{2^{3}x^{3}}=2x$ in the numerator, so we rationalize with
$ \displaystyle \frac{\sqrt[3]{2^{2}x}}{\sqrt[3]{y}}\color{red}{ \cdot\frac{\sqrt[3]{2x^{2}}}{\sqrt[3]{2x^{2}}} }\qquad$ (rationalize)
For the denominator, $\quad \sqrt[n]{a}\times\sqrt[n]{b}=\sqrt[n]{ab}$
$=\displaystyle \frac{2x}{\sqrt[3]{2x^{2}y}}$
