Example 10 Solve the system of equations
$$
\left\{\begin{array}{l}
x+y+\frac{9}{x}+\frac{4}{y}=10 \\
\left(x^{2}+9\right)\left(y^{2}+4\right)=24 x y .
\end{array}\right.
$$
The original system of equations can be transformed into
$$
\left\{\begin{array}{l}
\left(x+\frac{9}{x}\right)+\left(y+\frac{4}{y}\right)=10, \\
\left(x+\frac{9}{x}\right)\left(y+\frac{4}{y}\right)=24 .
\end{array}\right.
$$

By Vieta's Theorem, \( x+\frac{9}{x} \) and \( y+\frac{4}{y} \) are the roots of the equation \( t^{2}-10 t+24=0 \). Solving this equation, we get \( t=4 \) or \( t=6 \).
$$
\therefore\left\{\begin{array}{l}
x+\frac{9}{x}=4, \\
y+\frac{4}{y}=6
\end{array}\right. \text{ or } \left\{\begin{array}{l}
x+\frac{9}{x}=6, \\
y+\frac{4}{y}=4 .
\end{array}\right.
$$

Thus, (I) \(\left\{\begin{array}{l}x^{2}-4 x+9=0, \\ y^{2}-6 y+4=0\end{array}\right.\)
or (II) \(\left\{\begin{array}{l}x^{2}-6 x+9=0, \\ y^{2}-4 y+4=0 .\end{array}\right.\)
System (I) has no real solutions. Solving system (II), we get
$$
\left\{\begin{array}{l}
x=3, \\
y=2 .
\end{array}\right.
$$

Upon verification, \(\left\{\begin{array}{l}x=3, \\ y=2\end{array}\right.\) is a solution to the original system of equations.