Example 1 If real numbers $x, y$ satisfy
$$
\begin{array}{l}
1+\cos ^{2}(2 x+3 y-1) \\
=\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1},
\end{array}
$$

then the minimum value of $xy$ is
Analysis: This is an expression related to trigonometry, and solving it might involve the boundedness of trigonometric functions.
Given \(1+\cos ^{2}(2 x+3 y-1) \leqslant 2\), we predict
\[
\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1} \geqslant 2.
\]

In fact,
\[
\begin{array}{l}
\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1} \\
=\frac{(x-y)^{2}+2(x-y)+2}{x-y+1} \\
=\frac{(x-y+1)^{2}+1}{x-y+1} \\
=(x-y+1)+\frac{1}{x-y+1},
\end{array}
\]

Since \(\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1}>0\), the prediction is correct.
\[
\text { At this point }\left\{\begin{array}{l}
x-y+1=1, \\
2 x+3 y-1=k \pi
\end{array} \quad(k \in \mathbf{Z}) .\right.
\]

Solving these, we get \(x=y=\frac{k \pi+1}{5}\).
Therefore, \(x y=\left(\frac{k \pi+1}{5}\right)^{2} \geqslant \frac{1}{25}\). Hence, the answer is \(\frac{1}{25}\).