Example 2 (National I) If $\frac{\pi}{4}<x<\frac{\pi}{2}$, then the maximum value of the function $y=$ $\tan 2x \tan ^{3} x$ is
Given $\frac{\pi}{4}1$, we have $y=$ $\frac{2 \tan ^{4} x}{1-\tan ^{2} x}$. Let $t=1-\tan ^{2} x$, then $t \in(-\infty, 0)$. At this point, transforming it into a fractional function of $t$, and using the arithmetic mean-geometric mean inequality, we get
$$y=\frac{2(1-t)^{2}}{t}=-\left(-2 t+\frac{2}{-t}\right)-4 \leqslant-4-4=-8,$$

with equality holding only when $t=-1$, i.e., $\tan ^{2} x=2$. Therefore, the answer should be -8.