Answer
Domain: $(-\infty,-6)\cup(-6,\infty)$.
Work Step by Step
Domain restriction for logarithmic functions:
the argument of the logarithm must be positive.
$h(x)=\log_{3}(x+6)^{2}$
$(x+6)^{2}>0$
A square of a real number is nonnegative,
so the only problem is when the LHS is zero:
$x+6=0$
$x=-6\qquad $ ... $-6$ is the only number excluded
Domain: $(-\infty,-6)\cup(-6,\infty)$.
