Answer
$$\frac{y^3}{6x^4}$$
Work Step by Step
Using the rules of exponents, we find:
$$\frac{1}{\left(\frac{72x^3y^{-1}}{12x^{-1}y^2}\right)} \\ \frac{12x^{-1}y^2}{72x^3y^{-1}} \\ \frac{x^{-1}y^2}{6x^3y^{-1}} \\ \frac{y^2}{6x^4y^{-1}} \\ \frac{y^3}{6x^4}$$
