Example 3 Given $2 x>3 y>0$, find the minimum value of $\sqrt{2} x^{3}+\frac{3}{2 x y-3 y^{2}}$.

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Given that $2 x>3 y>0$, we have $2 x-3 y>0$. By the AM-GM inequality, we get
$$\begin{aligned}
2 x y-3 y^{2} & =y(2 x-3 y)=\frac{1}{3} \cdot 3 y(2 x-3 y) \\
& \leqslant \frac{1}{3} \cdot\left[\frac{3 y+(2 x-3 y)}{2}\right]^{2}=\frac{1}{3} x^{2}
\end{aligned}$$

Therefore,
$$\begin{aligned}
& \sqrt{2} x^{3}+\frac{3}{2 x y-3 y^{2}} \geqslant \sqrt{2} x^{3}+\frac{9}{x^{2}} \\
= & \frac{\sqrt{2}}{2} x^{3}+\frac{\sqrt{2}}{2} x^{3}+\frac{3}{x^{2}}+\frac{3}{x^{2}}+\frac{3}{x^{2}} \\
\geqslant & 5 \sqrt[5]{\left(\frac{\sqrt{2}}{2} x^{3}\right)^{2} \cdot\left(\frac{3}{x^{2}}\right)^{3}}=5 \sqrt[5]{\frac{27}{2}}
\end{aligned}$$

Equality holds if and only if $3 y=2 x-3 y, \frac{\sqrt{2}}{2} x^{3}=\frac{3}{x^{2}}$, i.e., when $x=18^{\frac{1}{10}}, y=\frac{1}{3} \cdot 18^{\frac{1}{10}}$.

Thus, the minimum value of $\sqrt{2} x^{3}+\frac{3}{2 x y-3 y^{2}}$ is $5 \sqrt[5]{\frac{27}{2}}$.