Answer
$\dfrac{1}{6y^{13}}$
Work Step by Step
Using the laws of exponents, the given expression, $
\dfrac{(-y^{-4})^2}{6(y^{-5})^{-1}}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{(-1)^2y^{-4(2)}}{6(y^{-5(-1)})}
\\\\=
\dfrac{1y^{-8}}{6y^{5}}
\\\\=
\dfrac{y^{-8-5}}{6}
\\\\=
\dfrac{y^{-13}}{6}
\\\\=
\dfrac{1}{6y^{13}}
.\end{array}