Example 7 Real numbers $a$ and $b$ satisfy the equation $\frac{a^{2} b^{2}}{a^{4}-2 b^{4}}=$
1. Then $\frac{a^{2}-b^{2}}{19 a^{2}+96 b^{2}}=$ $\qquad$
(1996, Beijing Junior High School Mathematics Competition)
Solution: The given equation can not only be seen as an equation of $a$, $b$, $a^{2}$, and $b^{2}$, but also as an equation of $\frac{a^{2}}{b^{2}}$ (or $\frac{b^{2}}{a^{2}}$).

In fact, from the given equation, we have
$$
a^{2} b^{2}=a^{4}-2 b^{4} \text {. }
$$

From the given equation, we know $b \neq 0$. Dividing both sides of the above equation by $b^{4}$, we get
$$
\frac{a^{2}}{b^{2}}=\frac{a^{4}}{b^{4}}-2 .
$$

Equation (1) is a quadratic equation in $\frac{a^{2}}{b^{2}}$:
$$
\left(\frac{a^{2}}{b^{2}}\right)^{2}-\frac{a^{2}}{b^{2}}-2=0 \text {. }
$$

Solving equation (2) yields $\frac{a^{2}}{t^{2}}=2$.
$$
\begin{array}{l}
\text { Therefore, } \frac{a^{2}-b^{2}}{19 a^{2}+96 b^{2}}=\frac{\frac{a^{2}}{b^{2}}-1}{19 \cdot \frac{a^{2}}{b^{2}}+96} \\
=\frac{2-1}{19 \times 2+96}=\frac{1}{134} . \\
\end{array}
$$