#### Answer

$\dfrac{\sqrt[3]{63xy^2}}{3y}$

#### Work Step by Step

Rationalizing the denominator of the given expression, $
\dfrac{\sqrt[3]{7x}}{\sqrt[3]{3y}}
,$ we find:
\begin{array}{l}\require{cancel}
\dfrac{\sqrt[3]{7x}}{\sqrt[3]{3y}}\cdot\dfrac{\sqrt[3]{9y^2}}{\sqrt[3]{9y^2}}
\\\\=
\dfrac{\sqrt[3]{63xy^2}}{\sqrt[3]{27y^3}}
\\\\=
\dfrac{\sqrt[3]{63xy^2}}{\sqrt[3]{(3y)^3}}
\\\\=
\dfrac{\sqrt[3]{63xy^2}}{3y}
.\end{array}
* Note that it is assumed that all variables represent positive numbers.