Answer
$(-\infty,-3)\cup(-3,3)\cup(3,\infty)$.
Work Step by Step
The domain requirement for $f(x)=\frac{x^2+7}{x^2-9}$ is $x^2-9\ne0$ which gives $x\ne\pm3$, in interval notation $(-\infty,-3)\cup(-3,3)\cup(3,\infty)$.
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