Answer
$\dfrac{q^{5/3}}{9p^{7/2}}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the laws of exponents to simplify the given expression, $
\left( \dfrac{p^{-1/4}q^{-3/2}}{ 3^{-1}p^{-2}q^{-2/3}} \right)^{-2}
.$
$\bf{\text{Solution Details:}}$
Using the extended Power Rule of the laws of exponents which states that $\left( \dfrac{x^my^n}{z^p} \right)^q=\dfrac{x^{mq}y^{nq}}{z^{pq}},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{p^{-\frac{1}{4}\cdot(-2)}q^{-\frac{3}{2}\cdot(-2)}}{3^{-1\cdot(-2)}p^{{-2}\cdot(-2)}q^{-\frac{2}{3}\cdot(-2)}}
\\\\=
\dfrac{p^{\frac{1}{2}}q^{3}}{3^{2}p^{4}q^{\frac{4}{3}}}
\\\\=
\dfrac{p^{\frac{1}{2}}q^{3}}{9p^{4}q^{\frac{4}{3}}}
.\end{array}
Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to
\begin{array}{l}\require{cancel}
\dfrac{p^{\frac{1}{2}-4}q^{3-\frac{4}{3}}}{9}
.\end{array}
Changing the exponents to similar fractions, the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{p^{\frac{1}{2}-\frac{8}{2}}q^{\frac{9}{3}-\frac{4}{3}}}{9}
\\\\=
\dfrac{p^{-\frac{7}{2}}q^{\frac{5}{3}}}{9}
.\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{q^{\frac{5}{3}}}{9p^{\frac{7}{2}}}
\\\\=
\dfrac{q^{5/3}}{9p^{7/2}}
.\end{array}
