Answer
$$-2$$
Work Step by Step
Recall that the definition $\sqrt{-a} = i\sqrt{a}$ must be applied BEFORE the other rules regarding radicals, therefore,
$\frac{\sqrt{-6}*\sqrt{-2}}{\sqrt 3}$
$\frac{i\sqrt{6}*i\sqrt{2}}{\sqrt 3}$
$(i^2)\frac{\sqrt{12}}{\sqrt 3}$
$(i^2)\sqrt {\frac{12}{3}}$
$(-1)(\sqrt{4})$
$$-2$$
