Inequality in disguise

We have to solve the equation $x^3+\frac{y^3}{3}+\frac{1}{9}=xy$ , it's kinda hard to solve by factoring, and sometimes it's better to invoke some inequalities in an equality! By AM-GM, $x^3+\frac{y^3}{3}+\frac{1}{9}\geq3(\ \frac{x^3y^3}{27})^\frac{1}{3}=xy$ , thus, there is only one element which is $(x,y)=(\frac{1}{9^\frac{1}{3}}, \frac{1}{3^\frac{1}{3}} )$ by setting $x^3=\frac{y^3}{3}=\frac{1}{9}$ which is the point where the function $x^3+\frac{y^3}{3}+\frac{1}{9}$ achieves its minimum value $xy$ .