Example 9 Given $x^{2}-y^{2}=16$. Find the minimum and maximum values of the function
$$
f(x, y)=\frac{1}{x^{2}}+\frac{y}{8 x}+1
$$
Solution: $x^{2}-y^{2}=16$ is a hyperbola with equal axes centered at the origin, and its parametric equations are
$$
x=4 \sec \theta, y=4 \tan \theta
$$
$(\theta$ is the parameter, and $\left.0 \neq \pm \frac{\pi}{2}\right)$.
Substituting them, we get
$$
\begin{array}{l}
f(x, y)=\frac{1}{x^{2}}+\frac{y}{8 x}+1 \\
=\frac{1}{16 \sec ^{2} \theta}+\frac{\tan \theta}{8 \sec \theta}+1=\frac{\cos ^{2} \theta}{16}+\frac{2 \sin \theta}{16}+1 \\
=-\frac{(\sin \theta-1)^{2}}{16}+\frac{9}{8} .
\end{array}
$$

Therefore, the maximum value of $f(x, y)$ is $\frac{9}{8}$, and the minimum value is $\frac{7}{8}$.