Answer
Domain at $$\{x\mid x\neq 9\}$$
Undefined point, or hole, at $$(9,6)$$
Work Step by Step
First off, the general function for this equation is $$f(x)=\frac{x-9}{\sqrt{x}-3}$$
Since the limit is when "x is approaching 9", which can be written as $$\underset{x\rightarrow 9}{\lim}$$ that would mean that there will be an undefined point at $(9,y)$ since $\sqrt{9}-3=3-3=0$ and anything with a denominator of $0$ is undefined.
So, since the function at the given limit is going to be undefined, then the first we can do is rationalize the numerator, which means you can multiply by $\sqrt{x}+3$, which is the conjugate of $\sqrt{x}-3$, which can be written as the following:
$$\begin{matrix}
\frac{x-9}{\sqrt{x}-3}\\\\
=\frac{\left(x-9\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\\\
=\frac{\left(x-9\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}\right)^2-3^2}\\\\
=\frac{\left(x-9\right)\left(\sqrt{x}+3\right)}{x-9}\\\\
=\frac{\left(1\right)\left(\sqrt{x}+3\right)}{1}\\\\
=\sqrt{x}+3
\end{matrix}$$
Now, onto the plug and play with the limitation of $$\sqrt{x}+3$$
Since the limit is when "x is approaching 9", which also means $x=9$, this means the following:
$$\begin{matrix}
_{x\rightarrow 9}^{\lim}\textrm{f(x)}&=\sqrt{(9)}+3\\
&=\sqrt{9}+3\\
&=3+3\\
&=6\\
\end{matrix}$$
Since the limit has an undefined point at $(9,y)$ and $_{x\rightarrow 9}^{\lim}\textrm{f(x)}=6$, which means that the (official) undefined point, or hole, at $$(9,6)$$ with the graph plotted down below.
So, since the domain is how the $x$-values range from left to right, and since $(9,6)$ is the undefined point, that would means that the domain would be all real numbers from $x\ge 0$ with the exception of at $x=9$, which can the written as the following:
$$\{x\mid x\neq 9\}$$
