Let $f(x)=\frac{3}{\cos x}+\frac{2}{\sin x}\left(0<x<\frac{\pi}{2}\right)$,
then $f^{\prime}(x)=\frac{3 \sin x}{\cos ^{2} x}-\frac{2 \cos x}{\sin ^{2} x}$.
Let $f^{\prime}(x)=0$, i.e., $\frac{3 \sin x}{\cos ^{2} x}-\frac{2 \cos x}{\sin ^{2} x}=0$.
Given $0<x<\frac{\pi}{2}$, we know $\sin x=\sqrt[2]{\frac{2}{3}} \cos x$.
Also, by $\sin ^{2} x+\cos ^{2} x=1$,
at this point, $x=\arctan \sqrt[3]{\frac{2}{3}}$ (list, determine the sign).
\begin{tabular}{cccc}
\hline$x$ & $\left(0, \arctan \sqrt[3]{\frac{2}{3}}\right)$ & $\arctan \sqrt[3]{\frac{2}{3}}$ & $\left(\arctan \sqrt[3]{\frac{2}{3}}, \frac{\pi}{2}\right)$ \\
\hline$f^{\prime}(x)$ & - & 0 & + \\
$f(x)$ & decreasing & minimum value & increasing \\
\hline
\end{tabular}

Since $f(x)$ has only one extremum point in $\left(0, \frac{\pi}{2}\right)$, it is also its minimum point.
Therefore, when $x=\arctan \sqrt[3]{\frac{2}{3}}$, $f(x)_{\text {minimum }}=\sqrt{(\sqrt[3]{9}+\sqrt[3]{4})^{3}}$.
Example 12 Find the minimum value of $\frac{3}{\cos x}+\frac{2}{\sin x}\left(0<x<\frac{\pi}{2}\right)$.