Chapter 5 - Exponents and Polynomials - 5.6 - Integral Exponents and Scientific Notation - Problem Set 5.6 - Page 224: 73

$\dfrac{y}{x^{2}}$

Using the laws of exponents, the given expression, $ \left( \dfrac{x^2}{y} \right)^{-1} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{x^{2(-1)}}{y^{1(-1)}} \\\\= \dfrac{x^{-2}}{y^{-1}} \\\\= \dfrac{y^{1}}{x^{2}} \\\\= \dfrac{y}{x^{2}} .\end{array}