Answer
$\frac{1}{9}$
Work Step by Step
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, $(-3)^{-2}=\frac{1}{(-3)^{2}}=\frac{1}{-3\times-3}=\frac{1}{9}$.