Answer
$-2$
Work Step by Step
Since $9=3^2$, the given expression is equivalent to:
$\log_{\frac{1}{3}}{(3^2)}$
Using the rule $a^{m} = \frac{1}{a^{-m}}, a\ne0$, the expression is equivalent to:
$\log_{\frac{1}{3}}{(3^2)}=\log_{\frac{1}{3}}{\left(\frac{1}{3^{-2}}\right)}$
Note that $\left(\frac{1}{3^{-2}}\right)=\left(\frac{1}{3}\right)^{-2}$.
Thus,
$\log_{\frac{1}{3}}{\left(\frac{1}{3^{-2}}\right)}=\log_{\frac{1}{3}}{\left(\left(\frac{1}{3}\right)^{-2}\right)}$
RECALL:
$\log_a{(a^n)} = n, a \gt 0, a \ne1$
Using the property above gives:
$\log_{\frac{1}{3}}{\left(\left(\frac{1}{3}\right)^{-2}\right)}=-2$
Thus, $\log_{\frac{1}{3}}{9} = -2$.