Example 1 If $\frac{2}{x}+\frac{3}{y}=2(x>0, y>0)$, find the minimum value of $x y$.


Since $x>0, y>0$, we get $2=\frac{2}{x}+\frac{3}{y} \geqslant 2 \sqrt{\frac{2}{x} \cdot \frac{3}{y}}$, thus $\sqrt{x y} \geqslant \sqrt{6}\left(\right.$ equality holds when $\frac{2}{x}=\frac{3}{y}$ and $\frac{2}{x}+\frac{3}{y}=2$, i.e., $x=2$ and $y=3$). Therefore, the minimum value of $x y$ is 6.