Answer
$(-\infty,-\sqrt 7)\cup(-\sqrt 7,\sqrt 7)\cup(\sqrt 7,\infty)$.
Work Step by Step
The domain requirement for $f(x)=\frac{1}{|x^2-7|}$ is $|x^2-7|\ne0$ or $x\ne\pm\sqrt 7$ , in interval notation $(-\infty,-\sqrt 7)\cup(-\sqrt 7,\sqrt 7)\cup(\sqrt 7,\infty)$.