Simplify the expression: $(\frac{2x^{-2}y}{x^2y^{-3}})^{-2} \div (\frac{xy^{-1}}{y})$

1. Problem Description

Simplify the expression:

$(\frac{2x^{-2}y}{x^2y^{-3}})^{-2} \div (\frac{xy^{-1}}{y})$

2. Solution Steps

First, simplify the fraction inside the first parenthesis:

$\frac{2x^{-2}y}{x^2y^{-3}} = 2 \cdot \frac{x^{-2}}{x^2} \cdot \frac{y}{y^{-3}} = 2 x^{-2-2} y^{1-(-3)} = 2x^{-4}y^4$

Then, raise this to the power of $-2$:

$(2x^{-4}y^4)^{-2} = 2^{-2} (x^{-4})^{-2} (y^4)^{-2} = \frac{1}{4} x^8 y^{-8}$

Next, simplify the fraction inside the second parenthesis:

$\frac{xy^{-1}}{y} = x \cdot \frac{y^{-1}}{y} = x y^{-1-1} = xy^{-2}$

The division becomes:

$(\frac{1}{4} x^8 y^{-8}) \div (xy^{-2}) = (\frac{1}{4} x^8 y^{-8}) \cdot \frac{1}{xy^{-2}} = \frac{1}{4} \cdot \frac{x^8}{x} \cdot \frac{y^{-8}}{y^{-2}} = \frac{1}{4} x^{8-1} y^{-8-(-2)} = \frac{1}{4} x^7 y^{-6}$

$a^{-n} = \frac{1}{a^n}$, so

$\frac{1}{4} x^7 y^{-6} = \frac{1}{4} x^7 \frac{1}{y^6} = \frac{x^7}{4y^6}$

3. Final Answer

$\frac{x^7}{4y^6}$