Chapter 5 - Section 5.7 - Negative Exponents and Scientific Notation - Exercise Set - Page 409: 58

$\dfrac{x^{2}}{16}$

Using the laws of exponents, the given expression, $ \left( 4x^{-1} \right)^{-2} ,$ simplifies to \begin{array}{l}\require{cancel} 4^{1(-2)}x^{-1(-2)} \\\\= 4^{-2}x^{2} \\\\= \dfrac{x^{2}}{4^{2}} \\\\= \dfrac{x^{2}}{16} .\end{array}