#### Answer

$\dfrac{\sqrt[5]{x^3}}{x}$

#### Work Step by Step

Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
(x^{-2/3})^{3/5}
\\\\=
x^{-\frac{2}{3}\cdot\frac{3}{5}}
\\\\=
x^{-\frac{2}{\cancel3}\cdot\frac{\cancel3}{5}}
\\\\=
x^{-\frac{2}{5}}
.\end{array}
Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
x^{-\frac{2}{5}}
\\\\=
\dfrac{1}{x^{\frac{2}{5}}}
\\\\=
\dfrac{1}{\sqrt[5]{x^2}}
.\end{array}
Multiplying by an expression equal to $1$ which will make the denominator a perfect power of the index results to
\begin{array}{l}\require{cancel}
\dfrac{1}{\sqrt[5]{x^2}}
\\\\=
\dfrac{1}{\sqrt[5]{x^2}}\cdot\frac{\sqrt[5]{x^3}}{\sqrt[5]{x^3}}
\\\\=
\dfrac{\sqrt[5]{x^3}}{\sqrt[5]{x^5}}
\\\\=
\dfrac{\sqrt[5]{x^3}}{x}
.\end{array}