Chapter 5 - Section 5.7 - Negative Exponents and Scientific Notation - Exercise Set - Page 408: 25

$\dfrac{y}{x^{8}}$

Using $x^{-a}=\dfrac{1}{x^a}$ and $\dfrac{1}{x^{-a}}=x^a,$ the given expression, $ \dfrac{x^{-8}}{y^{-1}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{y^{1}}{x^{8}} \\\\= \dfrac{y}{x^{8}} .\end{array}