$8 \cdot 37$ The number of elements in the point set $\left\{(x, y) \left\lvert\, \lg \left(x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}\right)=\lg x+\lg y\right.\right\}$ is
(A) 0.
(B) 1.
(C) 2.
(D) More than 2.
(China High School Mathematics League, 1990)
[Solution] From $\lg \left(x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}\right)=\lg x+\lg y$, we have
$$
\left\{\begin{array}{l}
x>0, \\
y>0, \\
x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}=x y .
\end{array}\right.
$$

By the AM-GM inequality,
$$
\begin{array}{l}
x^{3}+\frac{1}{3} y^{3}+\frac{1}{9} \geqslant 3 \sqrt[3]{\left(x^{3}\right)\left(\frac{1}{3} y^{3}\right)\left(\frac{1}{9}\right)}=x y, \\
\left\{\begin{array}{l}
x^{3}=\frac{1}{9}, \\
\frac{1}{3} y^{3}=\frac{1}{9},
\end{array}\right. \text { the equality holds. } \\
\end{array}
$$

The equality holds if and only if $\left\{\begin{array}{l}x^{3}=\frac{1}{9}, \\ \frac{1}{3} y^{3}=\frac{1}{9},\end{array}\right.$ Solving these two equations, we get $\quad x=\sqrt[3]{\frac{1}{9}}, \quad y=\sqrt[3]{\frac{1}{3}}$ (discarding the imaginary roots).
Thus, the set of points contains only one point $\left(\sqrt[3]{\frac{1}{9}}, \sqrt[3]{\frac{1}{3}}\right)$. Therefore, the answer is $(B)$.