Chapter 5 - Section 5.2 - Negative Exponents and Scientific Notation - Practice - Page 349: 10

$\frac{y^{2}}{81x^{6}}$

We know that $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}}=a^{n}$ (as long as a is a nonzero real number and n is an integer). Therefore, $(\frac{9x^{3}}{y})^{-2}=\frac{9^{-2}(x^{3})^{-2}}{y^{-2}}=\frac{9^{-2}x^{3\times-2}}{y^{-2}}=\frac{9^{-2}x^{-6}}{y^{-2}}=\frac{1}{9^{2}}\times\frac{1}{x^{6}}\times y^{2}=\frac{y^{2}}{81x^{6}}$.