Answer
$\dfrac{1}{9}$
Work Step by Step
RECALL:
(1) $a^m \cdot a^n = a^{m+n}$
(2) $a^{-m} = \dfrac{1}{a^m}, a \ne 0$
Use rule (1) above to obtain:
$=3^{-6+4}
\\=3^{-2}$
Use rule (2) above to obtain:
$=\dfrac{1}{3^2}
\\=\dfrac{1}{9}$