$$
\begin{array}{l}
\text { 3. } 1+\cos ^{2}(2 x+3 y-1) \\
=\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1}
\end{array}
$$

Then the minimum value of $xy$ is
3. $\frac{1}{25}$.

On one hand, $1 \leqslant 1+\cos ^{2}(2 x+3 y-1) \leqslant 2$.
On the other hand,
$$
\begin{array}{l}
\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1} \\
=\frac{(x-y)^{2}+2(x-y)+2}{x-y+1} \\
=(x-y+1)+\frac{1}{x-y+1} \geqslant 2 .
\end{array}
$$

Obviously, $x-y+1$ is a positive number.
Therefore, the original inequality holds if and only if $x-y+1=1$ and $2 x+3 y-1=k \pi(k \in \mathbf{Z})$, i.e., $x=y=\frac{k \pi+1}{5}$. Thus, $x y=\left(\frac{k \pi+1}{5}\right)^{2}$. Hence, the minimum value of $x y$ is $\frac{1}{25}$.