Answer
$\dfrac{2\sqrt[5]{m^4n^{2}}}{m^{2}n^{3}}$
Work Step by Step
Rationalizing the denominator of $
\sqrt[5]{\dfrac{32}{m^6n^{13}}}
$ results to
\begin{array}{l}
\sqrt[5]{\dfrac{32}{m^6n^{13}}}
\cdot
\sqrt[5]{\dfrac{m^4n^{2}}{m^4n^{2}}}
\\\\=
\sqrt[5]{\dfrac{32m^4n^{2}}{m^{10}n^{15}}}
\\\\=
\dfrac{2\sqrt[5]{m^4n^{2}}}{m^{2}n^{3}}
.\end{array}