Answer
$\dfrac{y}{18x}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the laws of exponents to simplify the given expression, $
(3x^2y^{-1})^{-2}(2x^{-3}y)^{-1}
.$
$\bf{\text{Solution Details:}}$
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}
(3^{-2}x^{2(-2)}y^{-1(-2)})(2^{-1}x^{-3(-1)}y^{-1})
\\\\=
(3^{-2}x^{-4}y^{2})(2^{-1}x^{3}y^{-1})
.\end{array}
Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(3^{-2})(2^{-1})x^{-4+3}y^{2+(-1)}
\\\\=
(3^{-2})(2^{-1})x^{-4+3}y^{2-1}
\\\\=
(3^{-2})(2^{-1})x^{-1}y^{1}
\\\\=
(3^{-2})(2^{-1})x^{-1}y
.\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{y}{(3^{2})(2^{1})x^{1}}
\\\\=
\dfrac{y}{(9)(2)x}
\\\\=
\dfrac{y}{18x}
.\end{array}