The implicit function $z=z(x, y)$ (satisfying the conditions of the implicit function theorem), and considering the target function $f(x, y, z)=x y z(x, y)=F(x, y)$ as a composite function of $f$ and $z=z(x, y)$. This way, we can apply the sufficient conditions for extrema to make a judgment. For this purpose, the following calculations are made:



保留源文本的换行和格式，直接输出翻译结果如下：

The implicit function $z=z(x, y)$ (satisfying the conditions of the implicit function theorem), and considering the target function $f(x, y, z)=x y z(x, y)=F(x, y)$ as a composite function of $f$ and $z=z(x, y)$. This way, we can apply the sufficient conditions for extrema to make a judgment. For this purpose, the following calculations are made:
Let $G(x, y, z)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{a}$, then

thus
$$G_{x}=-\frac{1}{x^{2}}, G_{y}=-\frac{1}{y^{2}}, G_{z}=-\frac{1}{z^{2}}, z_{x}=-\frac{G_{x}}{G_{z}}=-\frac{z^{2}}{x^{2}}, z_{y}=-\frac{z^{2}}{y^{2}}$$
$$\begin{array}{c}
F_{x}=y z+x y z_{x}=y z-\frac{y z^{2}}{x}, F_{y}=x z+x y z_{y}=x z-\frac{x z^{2}}{y} \\
F_{x x}=y z_{x}+y z_{x}+x y z_{x x}=\frac{2 y z^{2}}{x^{3}} \\
F_{x y}=z+y z_{y}+x y z_{x y}=z-\frac{z^{2}}{y}-\frac{z^{2}}{x}+\frac{2 z^{3}}{x y} \\
F_{y y}=x z_{y}+x z_{y}+x y z_{y y}=\frac{2 x z^{3}}{y^{3}}
\end{array}$$

When $x=y=z=3 a$,
$$\begin{array}{c}
A=F_{x x}=6 a=F_{y y}=C, \quad B=F_{x y}=3 a, \\
A C-B^{2}=36 a^{2}-9 a^{2}=27 a^{2}>0, \text { and } A=6 a>0
\end{array}$$

From this, we can see that the required stable point is a minimum point. Therefore, the objective function $u=x y z$ under the constraint $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{a}$ achieves a minimum value of $27 a^{3}$ at the point $(3 a, 3 a, 3 a)$.