The problem is to simplify the expression $\frac{[(2x^2y^3)^2]^{-2}}{3x^{-2}y}$ and express the answer in simplest positive indicial form.

1. Problem Description

The problem is to simplify the expression $\frac{[(2x^2y^3)^2]^{-2}}{3x^{-2}y}$ and express the answer in simplest positive indicial form.

2. Solution Steps

First, simplify the numerator:

$[(2x^2y^3)^2]^{-2} = (2x^2y^3)^{-4} = 2^{-4}x^{-8}y^{-12} = \frac{1}{2^4x^8y^{12}} = \frac{1}{16x^8y^{12}}$

Next, rewrite the original expression:

$\frac{[(2x^2y^3)^2]^{-2}}{3x^{-2}y} = \frac{1}{16x^8y^{12}} \cdot \frac{1}{3x^{-2}y} = \frac{1}{16x^8y^{12} \cdot 3x^{-2}y} = \frac{1}{48x^{8-2}y^{12+1}} = \frac{1}{48x^6y^{13}}$

We have:

$x^a \cdot x^b = x^{a+b}$

$(x^a)^b = x^{ab}$

3. Final Answer

$\frac{1}{48x^6y^{13}}$