Answer
$\dfrac{81y^{8}}{x^{12}}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
\left( \dfrac{1}{3}x^3y^{-2} \right)^{-4}
,$ use the laws of exponents.
$\bf{\text{Solution Details:}}$
Using the extended Power Rule of the laws of exponents, which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\left( \dfrac{1}{3}x^3y^{-2} \right)^{-4}
\\\\=
\left( \dfrac{1}{3}\right)^{-4}x^{3(-4)}y^{-2(-4)}
\\\\=
\left( \dfrac{1}{3}\right)^{-4}x^{-12}y^{8}
.\end{array}
Using the Power of a Quotient Rule of the laws of exponents, which is given by $\left( \dfrac{x^m}{y^n} \right)^p=\dfrac{x^{mp}}{y^{np}},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\left( \dfrac{1}{3}\right)^{-4}x^{-12}y^{8}
\\\\=
\dfrac{1^{-4}}{3^{-4}}x^{-12}y^{8}
\\\\=
\dfrac{1^{-4}\cdot x^{-12}y^{8}}{3^{-4}}
.\end{array}
Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{1^{-4}\cdot x^{-12}y^{8}}{3^{-4}}
\\\\=
\dfrac{3^{4}\cdot y^{8}}{1^{4}x^{12}}
\\\\=
\dfrac{81y^{8}}{x^{12}}
.\end{array}
