Given $x^{2}-9 y^{2}-4 x+18 y-9=$
0. Find the extremum of the function
$$
f(x, y)=\frac{5 x^{2}+3 x y-23 x-6 y+22}{x^{2}-4 x+4}
$$
Solving: $x^{2}-9 y^{2}-4 x+18 y-9=0$ can be transformed into $\frac{(x-2)^{2}}{4}-\frac{(y-1)^{2}}{\frac{4}{9}}=1$, its parametric equations are
$$
x=2+2 \sec \theta, y=1+\frac{2}{3} \tan \theta
$$
( $\theta$ is the parameter, $0<\theta<\frac{\pi}{2}$ ).
Substituting into $f(x, y)$ gives
$$
\begin{aligned}
f(x, y) & =\frac{5(x-2)^{2}+3(x-2)(y-1)-4}{(x-2)^{2}} \\
& =\frac{5 \times 4 \sec ^{2} \theta+3 \times 2 \sec \theta \times \frac{2}{3} \tan \theta-4}{4 \sec ^{2} \theta} \\
& =5+\sin \theta-\cos ^{2} \theta \\
& =\sin ^{2} \theta+\sin \theta+4 \\
& =\left(\sin \theta+\frac{1}{2}\right)^{2}+\frac{15}{4} .
\end{aligned}
$$

Therefore, the maximum and minimum values of $f(x, y)$ are 6 and $\frac{15}{4}$, respectively.