Answer
The vertical asymptote will be $x=-3$; there will be a hole at $x=3$.
Work Step by Step
Write $g\left( x \right)$ in the form of $\frac{p\left( x \right)}{q\left( x \right)}$.
$\begin{align}
& g\left( x \right)=\frac{x-3}{{{x}^{2}}-9} \\
& =\frac{p\left( x \right)}{q\left( x \right)}
\end{align}$
So, $p\left( x \right)=x-3$ and $q\left( x \right)={{x}^{2}}-9$.
For, the vertical asymptotes, put $q\left( x \right)=0$:
$\begin{align}
& q\left( x \right)={{x}^{2}}-9=0 \\
& {{x}^{2}}-9=0 \\
& \left( x-3 \right)\left( x+3 \right)=0
\end{align}$
Thus, the vertical asymptotes are $x=3,\text{ or }x=-3$
But, after rearranging, $x-3$ from the denominator will be eliminated.
So, the vertical asymptote will be $x=-3$.
To find the holes, observe the expression:
$\begin{align}
& g\left( x \right)=\frac{x-3}{{{x}^{2}}-9} \\
& =\frac{x-3}{\left( x-3 \right)\left( x+3 \right)} \\
& =\frac{1}{x+3}
\end{align}$
Here, $x-3$ will be eliminated; when x approaches 3, the polynomial approaches to $\frac{1}{6}$ , but at the point $x=3$ the polynomial will be discontinuous. So, there will be a hole at $x=3$.
Therefore, the vertical asymptote will be $x=-3$; there will be a hole at $x=3$.
Hence, the vertical asymptote will be $x=-3$; there will be a hole at $x=3$.
