Example 10 Let $x>y>0, xy=1$, find the minimum value of $\frac{3x^3+125y^3}{x-y}$.

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Example 10 Let $x>y>0, xy=1$, find the minimum value of $\frac{3x^3+125y^3}{x-y}$.
Let the required minimum value (target) be a positive number $t$, then $\frac{3 x^{3}+125 y^{3}}{x-y} \geqslant t$, which means $3 x^{3}+125 y^{3}+y t \geqslant t x$. By $x y=1$ and the AM-GM inequality, we can get
$$
\begin{array}{l}
3 x^{3}+125 y^{3}+y t=x^{3}+x^{3}+x^{3}+125 y^{3}+y t \\
\geqslant 5 \sqrt[5]{x^{9} \cdot 125 y^{4} t}=5 \sqrt[5]{125 x^{5} t}=5 \sqrt[5]{125 t x},
\end{array}
$$

Let $5 \sqrt[5]{125 t x}=t x$, solving this gives $t=25$. Note that the above inequality holds with equality if and only if $x^{3}=125 y^{3}=y t$,
Therefore, it can be solved that when $x=\sqrt{5}, y=\frac{\sqrt{5}}{5}$, $\frac{3 x^{3}+125 y^{3}}{x-y}$ reaches its minimum value of 25.