5. (1990 National High School Mathematics Competition) The number of elements in the 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
5. Choose B. Reason: From $\lg \left(x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}\right)=\lg x+\lg$,

we get $\left\{\begin{array}{l}x>0, \\ y>0, \\ x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}=x y .\end{array}\right.$
Therefore, the problem of the number of elements in the set of points is equivalent to the number of ordered pairs $(x, y)$ that satisfy (1).
Notice that, $x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}=x y=3 \sqrt[3]{x^{3} \cdot \frac{1}{3} y^{3} \cdot \frac{1}{9}}$,
so by using the arithmetic and geometric mean inequality, we have
$x^{3}+\frac{1}{3} y^{3}+\frac{1}{9} \geqslant 3 \sqrt[3]{x^{3} \cdot \frac{1}{3} y^{3} \cdot \frac{1}{9}}=x y$, where equality holds if and only if $x^{3}=\frac{1}{3} y^{3}=\frac{1}{9}$.
Thus, the problem is reduced to
finding the number of real solutions to the system of equations $\left\{\begin{array}{l}x^{3}=\frac{1}{9}, \\ \frac{1}{3} y^{3}=\frac{1}{9}\end{array}\right.$
It is not difficult to find that the solution to (2) is $x=\sqrt[3]{\frac{1}{9}}, y=\sqrt[3]{\frac{1}{3}}$.
Hence, the set of points has only one element.