Example 11 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" Invitational Competition Question)
Solution: Since $(x-y)^{2}+\left(x+\frac{1}{y}+1\right)^{2}$
$$
=(x-y)^{2}+\left[(-x-1)-\frac{1}{y}\right]^{2},
$$

the above expression can be seen as the square of the distance between the point $(x,-x-1)$ on the line $y=-x-1$ and the point $\left(y, \frac{1}{y}\right)$ on the hyperbola $x y=1$, as shown in Figure 4-1.

Therefore, the minimum value sought is the square of the shortest distance between the line and the hyperbola. From the graph of the function, it is known that the coordinates of points $A, B, C$ are $(1,1)$, $(-1,1)$, and $\left(-\frac{1}{2},-\frac{1}{2}\right)$, respectively. Thus, the sought minimum value is $|BC|$. Also, $|BC|=\sqrt{2}-\frac{\sqrt{2}}{2}=\frac{\sqrt{3}}{2}$, and $|BC|^{2}=\frac{1}{2}$. Hence, $f(x, y)_{\text{min}}=\frac{1}{2}$ is the answer.

Note: By $(x-y)^{2}+\left(x+\frac{1}{y}+1\right)^{2}=(x-y)^{2}+\left[(x+1)-\left(-\frac{1}{y}\right)\right]^{2}$, we consider the square of the distance between the point $(y,-\frac{1}{y})$ on the hyperbola $x y=-1$ and the line $y=x+1$.