Answer
$\dfrac{3y}{\sqrt[3]{33y^2}}$
Work Step by Step
Multiplying both the numerator and the denominator by $
3y^2
$, then the rationalized-numerator form of the given expression, $
\sqrt[3]{\dfrac{9y}{11}}
,$ is
\begin{array}{l}\require{cancel}
\sqrt[3]{\dfrac{9y}{11}\cdot\dfrac{3y^2}{3y^2}}
\\\\=
\sqrt[3]{\dfrac{27y^3}{33y^2}}
\\\\=
\dfrac{\sqrt[3]{27y^3}}{\sqrt[3]{33y^2}}
\\\\=
\dfrac{\sqrt[3]{(3y)^3}}{\sqrt[3]{33y^2}}
\\\\=
\dfrac{3y}{\sqrt[3]{33y^2}}
.\end{array}
