Answer
$\dfrac{5\sqrt[3]{9y^2}}{3y}$
Work Step by Step
Rationalizing the denominator of $
\dfrac{5}{\sqrt[3]{3y}}
$ results to
\begin{array}{l}
\dfrac{5}{\sqrt[3]{3y}}
\cdot
\dfrac{\sqrt[3]{9y^2}}{\sqrt[3]{9y^2}}
\\\\=
\dfrac{5\sqrt[3]{9y^2}}{\sqrt[3]{27y^3}}
\\\\=
\dfrac{5\sqrt[3]{9y^2}}{3y}
.\end{array}