Example 1 Find the minimum value of the function with real variables $x$ and $y$
$$
u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}
$$

(2nd "Hope Cup" National Mathematics Invitational Competition)
Explanation: The original formula is transformed to
$$
u(x, y)=\left(\frac{9}{x}+\sqrt{2-y^{2}}\right)^{2}+(x-y)^{2}-2 \text {. }
$$

Consider the points $P_{1}\left(x, \frac{9}{x}\right), P_{2}\left(y,-\sqrt{2-y^{2}}\right)$. When $x \in \mathbf{R}(x \neq 0)$, point $P_{1}$ lies on a hyperbola with the coordinate axes as asymptotes, and point $P_{2}$ lies on the semicircle $x^{2}+y^{2}=2$ $(y \leqslant 0)$, then
$$
\left|P_{1} P_{2}\right| \geqslant 3 \sqrt{2}-\sqrt{2}=2 \sqrt{2} \text {. }
$$

Therefore, $u(x, y)=\left|P_{1} P_{2}\right|^{2}-2 \geqslant 6$.
When and only when $x=-3, y=-1$, $u(x, y)=6$. Thus, $u(x, y)_{\text {min }}=6$.

Note: The key to this solution is to ingeniously convert the extremum of a bivariate function into the shortest distance between a point on a hyperbola and a point on a semicircle.