Answer
$\frac{(xy^{-1}z^3)^2}{x^2yz^2} = \frac{z^4}{y^3} $
Work Step by Step
By the formulas given on page eight and nine,
$\frac{(xy^{-1}z^3)^2}{x^2yz^2} = \frac{x^2y^{-1\times2}z^{3\times2}}{x^2yz^2} = \frac{x^2y^{-2}z^6}{x^2yz^2} =x^{2-2}y^{(-2)-1}z^{6-2} = x^0 y^{-3} z^4 = y^{-3} z^4 = \frac{z^4}{y^3} $.
