## Precalculus (10th Edition)

All real numbers except $-3$ and $3$.
Note that $x^2-9=(x+3)(x-3).$ The denominator cannot be $0$. Hence, $x-3\ne0\longrightarrow x\ne 3$ $x+3\ne0\longrightarrow x\ne-3$ Thus, the domain is that set of all real numbers except $-3$ and $3$.