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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Explanation: Since $x>y>0$, we can set $x=y+\Delta y$ $(\Delta y>0)$, then
$$
\frac{x^{2}+y^{2}}{x-y}=\frac{(x-y)^{2}+2 x y}{x-y}=\frac{(\Delta y)^{2}+2}{\Delta y} \geqslant 2 \sqrt{2} \text{. }
$$

Equality holds if and only if $\Delta y=\sqrt{2}$, i.e.,
$$
x=\frac{\sqrt{2}+\sqrt{6}}{2}, y=\frac{\sqrt{6}-\sqrt{2}}{2}
$$

In this case, the minimum value of $\frac{x^{2}+y^{2}}{x-y}$ is $2 \sqrt{2}$.