Example 6 Let $x y=1$, and $x>y>0$, find the minimum value of $\frac{x^{2}+y^{2}}{x-y}$.

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Given $x>y>0$, we can set $x=y+\Delta y(\Delta y>0)$, then $\frac{x^{2}+y^{2}}{x-y}=\frac{(x-y)^{2}+2 x y}{x-y}=\frac{(\Delta y)^{2}+2}{\Delta y} \geqslant 2 \sqrt{2}$,
the equality holds if and only if $\Delta y=\sqrt{2}$, i.e., $x=\frac{\sqrt{2}+\sqrt{6}}{2}, y=\frac{\sqrt{6}-\sqrt{2}}{2}$.