Answer
$\dfrac{1}{16}$
Work Step by Step
Using $x^{-a}=\dfrac{1}{x^a}$ and $\dfrac{1}{x^{-a}}=x^a,$ the given expression, $
\dfrac{4^{-3}}{2^{-2}}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{2^{2}}{4^{3}}
\\\\=
\dfrac{4}{64}
\\\\=
\dfrac{\cancel{4}}{\cancel{4}\cdot16}
\\\\=
\dfrac{1}{16}
.\end{array}