(6*x^3*y^-7 divided by 9xy)^-1 ????
\(\left(\dfrac{6x^3y^{-7}}{9xy}\right)^{-1}\\ = \dfrac{9xy}{6x^3y^{-7}}\\ = \dfrac{9xy^8}{6x^3}\\ =\dfrac{3y^8}{2x^2}\)
