How many ordered pairs satisfy $\log(x^3+\frac{1}{3}y^3+\frac{1}{9})=\log x+\log y$? How many ordered pairs satisfy $\log(x^3+\frac{1}{3}y^3+\frac{1}{9})=\log x+\log y$?

It simplifies to $x^3+\frac{1}{3}y^3+\frac{1}{9}=xy$
I dont know how to proceed further.
 A: $$\log\left(x^3+\frac{1}{3}y^3+\frac{1}{9}\right)=\log x+\log y\tag1$$
First of all, we have to have
$$x^3+\frac{1}{3}y^3+\frac{1}{9}\gt 0\quad\text{and}\quad x\gt 0\quad\text{and}\quad y\gt 0\tag2$$
Under $(2)$, we have
$$(1)\implies x^3+\frac{1}{3}y^3+\frac{1}{9}-xy=0\tag3$$
As lab bhattacharjee suggests in the comments, the LHS of $(3)$ is of the form $$A^3+B^3+C^3-3ABC$$
Letting 
$$a=x,\qquad b=\frac{y}{\sqrt[3]3},\qquad c=\frac{1}{\sqrt[3]{9}}$$
we have
$$\begin{align}(3)&\implies x^3+\left(\frac{y}{\sqrt[3]3}\right)^3+\left(\frac{1}{\sqrt[3]{9}}\right)^3-3\cdot x\cdot\frac{y}{\sqrt[3]{3}}\cdot\frac{1}{\sqrt[3]{9}}=0
\\\\&\implies a^3+b^3+c^3-3abc=0
\\\\&\implies (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0
\\\\&\implies (a+b+c)\times \frac 12((a-b)^2+(b-c)^2+(c-a)^2)=0
\\\\&\implies a+b+c=0\quad\text{or}\quad a=b=c
\\\\&\implies x+\frac{y}{\sqrt[3]3}+\frac{1}{\sqrt[3]{9}}=0\qquad\text{or}\qquad (x,y)=\left(\frac{1}{\sqrt[3]{9}},\frac{1}{\sqrt[3]3}\right)\end{align}$$
Since $x\gt 0$ and $y\gt 0$ from $(2)$, we see that 
$$x+\frac{y}{\sqrt[3]3}+\frac{1}{\sqrt[3]{9}}=0$$
does not hold.
Since $(x,y)=\left(\frac{1}{\sqrt[3]{9}},\frac{1}{\sqrt[3]3}\right)$ satisfies $(2)$, we see that the only solution for $(1)$ is
$$x=\frac{1}{\sqrt[3]{9}},\qquad y=\frac{1}{\sqrt[3]3}$$
