2. Solve the system of equations $\left\{\begin{array}{l}3 x-y-3 x y=-1, \\ 9 x^{2} y^{2}+9 x^{2}+y^{2}-6 x y=13 .\end{array}\right.$
Answer: $\left(-\frac{2}{3} ; 1\right),(1 ; 1),\left(-\frac{1}{3} ;-3\right),\left(-\frac{1}{3} ; 2\right)$.

Solution. Rewrite the system as $\left\{\begin{array}{l}(3 x-y)-3 x y=-1, \\ (3 x y)^{2}+(3 x-y)^{2}=13\end{array}\right.$, then introduce new variables: $u=3 x-y, v=3 x y$. The system becomes $\left\{\begin{array}{l}u-v=-1, \\ v^{2}+u^{2}=13 .\end{array}\right.$ From the first equation, $v=1+u$. Substituting this into the second equation, we get $(u+1)^{2}+u^{2}=13, u^{2}+u-6=0$, from which it follows that $u=-3$ (and then $v=-2$) or $u=2$ (and then $v=3$).

If $u=-3, v=-2$, then $\left\{\begin{array}{l}3 x-y=-3, \\ 3 x y=-2\end{array} \Leftrightarrow\left\{\begin{array}{l}y=3 x+3, \\ 9 x^{2}+9 x+2=0 .\end{array}\right.\right.$ From this, we get that either $x=-\frac{1}{3}$, $y=2$, or $x=-\frac{2}{3}, y=1$.

If $u=2, v=3$, then $\left\{\begin{array}{l}3 x-y=2, \\ 3 x y=3\end{array} \Leftrightarrow\left\{\begin{array}{l}y=3 x-2, \\ 3 x^{2}-2 x-1=0 .\end{array}\right.\right.$ From this, we get that either $x=-\frac{1}{3}, y=-3$, or $x=1, y=1$.