One vertex of the hyperbola is $\left(\mathrm{3,0}\right).$

So, assuming the equation of hyperbola as $\frac{x^2}{9}-\frac{y^2}{b^2}=1$
and passing through $\left(3\sqrt{2},2\right)$, we get, 
$$\begin{gathered} \Rightarrow \frac{9\left(2\right)}{9}-\frac{4}{b^2}=1 \\ \Rightarrow 1=\frac{4}{b^2}\Rightarrow b^2=4 \end{gathered}$$
$\Rightarrow e^2=1+\frac{b^2}{a^2}=1+\frac{4}{9}=\frac{13}{9}$
$\Rightarrow \frac{e^2+1}{e^2-1}=\frac{\frac{13}{9}+1}{\frac{13}{9}-1}=\frac{22}{4}=\frac{11}{2}=5.5$