Chapter 5 - Exponents and Polynomials - 5.6 - Integral Exponents and Scientific Notation - Problem Set 5.6 - Page 223: 29

$144$

Using the laws of exponents, the given expression, $ \left( \dfrac{4^{-1}}{3} \right)^{-2} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{4^{-1(-2)}}{3^{1(-2)}} \\\\= \dfrac{4^{2}}{3^{-2}} \\\\= 4^{2}\cdot3^2 \\\\= 16\cdot9 \\\\= 144 .\end{array}