Answer
Set $\ \ g(3)=6$
Work Step by Step
We see that $g(3) $ is not defined, and g is continuous everywhere else.
We check if $\displaystyle \lim_{x\rightarrow 3}g(x)=L$ exists, and if it does, we can define the continuous extension of $g$ at $x=3$ as
$G(x)=\left\{\begin{array}{ll}
g(x) , & \text{if }x\neq 3\\
L, & \text{if }x=3
\end{array}\right.$
$L= \displaystyle \lim_{x\rightarrow 3}g(x)= \displaystyle \lim_{x\rightarrow 3}\frac{x^{2}-9}{x-3}\quad $...recognize a difference of squares,
$= \displaystyle \lim_{x\rightarrow 3}\frac{(x+3)(x-3)}{(x-3)}\quad $...cancel common term
$= \displaystyle \lim_{x\rightarrow 3}(x+3)$
... evaluate directly,
$L=6$
The limit exists at $x=3,$ so, if we redefine $g$,
$g(x)=\left\{\begin{array}{ll}
\frac{x^{2}-9}{x-3} , & \text{if }x\neq 3\\
6, & \text{if }x=3
\end{array}\right.$
it becomes continuous.