10. Solve the system of equations in the set of real numbers $\left\{\begin{array}{l}x^{2}+3 x y=18, \\ x y+3 y^{2}=6 .\end{array}\right.$
10. The original system of equations can be transformed into $\left\{\begin{array}{l}x^{2}+3 y x=18, \\ y x+3 y^{2}=6 .\end{array}\right.$

From (2), we know $y \neq 0$ and $x=\frac{6-3 y^{2}}{y}$.
Substituting (3) into (1) gives $x^{2}=9 y^{2}$, thus $\left(\frac{6-3 y^{2}}{y}\right)^{2}=9 y^{2}$, simplifying to $y^{2}=1$, yielding $y= \pm 1$. Substituting $y=1$ into (3) gives $x_{1}=3$; substituting $y=-1$ into (3) gives $x_{2}=-3$. Therefore, the solutions $(x, y)$ to the original system of equations are $(3,1),(-3,-1)$.