Thank you all for your help today! Here's another that's been stumping me:
(2X¯¹Y²)(4X²Y¯³)¯²
(12X¯²Y¯²)¯¹
$\displaystyle \frac{(2x^{-1}y^2)(4x^2y^{-3})^{-2}}{(12x^{-2}y^{-2})^{-1}}=\frac{2x^{-1}y^2(2^2)^{-2}(x^2)^{-2}(y^{-3})^{-2}}{(2^2\cdot 3)^{-1}(x^{-2})^{-1}(y^{-2})^{-1}}=\frac{2x^{-1}y^22^{-4}x^{-4}y^6}{2^{-2}3^{-1}x^2y^2}=$
$\displaystyle =\frac{2^{-3}x^{-5}y^8}{2^{-2}3^{-1}x^2y^2}=\frac{3}{2}x^{-7}y^6$