Answer
$\displaystyle \frac{x^{12}}{9y^{8}}$
Work Step by Step
... apply: $(abc)^{n}=a^{n}b^{n}c^{n}$
$=(-1)^{-2}(3)^{-2}(x^{-6})^{-2}(y^{4})^{-2}\qquad $ ... apply $(a^{m})^{n}=a^{mn}$, and $a^{-n}=\displaystyle \frac{1}{a^{n}}$
$=[(-1)^{2}]^{-1}(\displaystyle \frac{1}{3^{2}})x^{12}y^{-8}$
$=\displaystyle \frac{1^{-1}x^{12}}{3^{2}y^{8}}$
$=\displaystyle \frac{x^{12}}{9y^{8}}$
