Domain

$\dfrac{\sqrt{x^2-2}}{x^2-4}$

For this expression to be defined, we have two conditions:

Condition 1 : Whatever that is in the square root must be non-negative

$x^2 - 2 \geq 0\implies x^2 \geq 2 \implies x \leq -\sqrt{2},x \geq \sqrt{2}$

Condition 2 : The denominator of a fraction cannot be $0$

$x^2-4 \neq 0 \implies x^2 \neq 4 \implies x \neq \pm 2$

Combine these two conditions, and we have:

$x \leq -\sqrt{2},x \geq \sqrt{2},x \neq \pm 2$

In set notation: $\left(-\infty,-2\right)\cup\left(-2,-\sqrt{2}\right]\cup\left[\sqrt{2},2\right)\cup\left(2,\infty\right)$

Another way to write it: $\boxed{\mathbb R \backslash\left(-\sqrt{2},\sqrt{2}\right)\cup\{-2,2\}}$