Answer
$\dfrac{9}{16}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
\left( -\dfrac{64}{27} \right)^{-2/3}
,$ use the laws of exponents.
$\bf{\text{Solution Details:}}$
Using the extended Power Rule of the laws of exponents which states that $\left( \dfrac{y^n}{z^p} \right)^q=\dfrac{y^{nq}}{z^{pq}},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\left( \dfrac{-64}{27} \right)^{-2/3}
\\\\=
\dfrac{(-64)^{-2/3}}{(27)^{-2/3}}
.\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}
\dfrac{(27)^{2/3}}{(-64)^{2/3}}
.\end{array}
Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{(\sqrt[3]{27})^{2}}{(\sqrt[3]{-64})^{2}}
\\\\=
\dfrac{(\sqrt[3]{3^3})^2}{\sqrt[3]{(-4)^3})^2}
\\\\=
\dfrac{(3)^{2}}{(-4)^{2}}
\\\\=
\dfrac{9}{16}
.\end{array}
