Answer
The domain is all real numbers except $-3$ and $3$.
Work Step by Step
We will factor the denominator to obtain:
$f(x)=\dfrac{x}{x^2-9} =\dfrac{x}{(x-3)(x+3)} $
The function is undefined when the denominator becomes $0$.
This means that the value of $x$ can be any real number except those values that will make the denominator equal to zero.
Thus, $x+3\ne 0 \implies x \neq -3$ and $x-3\ne0 \implies x\ne3$.
So, the domain is all real numbers except $-3$ and $3$.