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

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Solution: Since $x>y>0$, we can set $x=y+\Delta y(\Delta y>0)$,

then $\frac{x^{2}+y^{2}}{x-y}=\frac{(x-y)^{2}+2xy}{x-y}=\frac{(\Delta y)^{2}+2}{\Delta y} \geqslant$ $2 \sqrt{2}$.

The equality holds if and only if $\Delta y=\sqrt{2}$, i.e., $x=\frac{\sqrt{2}+\sqrt{6}}{2}, y=$ $\frac{\sqrt{6}-\sqrt{2}}{2}$.
Therefore, the minimum value of $\frac{x^{2}+y^{2}}{x-y}$ is $2 \sqrt{2}$.