Answer
$(-\infty,-3)\cup(-3,3)\cup(3,\infty)$
Work Step by Step
The denominator of the given function, $
f(x)=\dfrac{x^2+7}{x^2-9}
,$ cannot be $0$. Hence,
\begin{array}{l}\require{cancel}
x^2-9\ne0
\\\\
x^2\ne9
\\\\
x\ne\pm\sqrt{9}
\\\\
x\ne\pm3
.\end{array}
Hence, the domain is $
(-\infty,-3)\cup(-3,3)\cup(3,\infty)
.$