Example 13 Find the minimum value of the bivariate function $f(x, y)=(x-y)^{2}+\left(x+\frac{1}{y}+1\right)^{2}$.
(1998 "Hope Cup" Competition Question)
Solve: Since $f(x, y)=(x-y)^{2}+\left(-x-1-\frac{1}{y}\right)^{2}$ can be seen as the squared distance between a point $(x, -x-1)$ on the line $x+y+1=0$ and a point $\left(y, \frac{1}{y}\right)$ on the hyperbola $xy=1$. From the graph, we can see that the required minimum value is $\frac{1}{2}$.