Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
xy\left( {x + 3} \right)\left( {y + 2} \right) = - 6\\
{x^2} + {y^2} + 3x + 2y = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {{x^2} + 3x} \right).\left( {{y^2} + 2y} \right) = - 6\\
\left( {{x^2} + 3x} \right) + \left( {{y^2} + 2y} \right) = 1
\end{array} \right.\\
Đặt:\left\{ \begin{array}{l}
{x^2} + 3x = a\\
{y^2} + 2y = b
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
ab = - 6\\
a + b = 1
\end{array} \right.\\
\Rightarrow a.\left( {1 - a} \right) = - 6\\
\Rightarrow {a^2} - a - 6 = 0\\
\Rightarrow \left( {a - 3} \right)\left( {a + 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
a = 3 \Rightarrow b = - 2\\
a = - 2 \Rightarrow b = 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^2} + 3x = 3\\
{y^2} + 2y = - 2\left( {vn} \right)
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} + 3x = - 2\\
{y^2} + 2y = 3
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} + 3x + 2 = 0\\
{y^2} + 2y - 3 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = - 1/x = - 2\\
y = 1/y = - 3
\end{array} \right.\\
Vậy\,\left( {x;y} \right) \in \left\{ {\left( { - 1;1} \right);\left( { - 1; - 3} \right);\left( { - 2;1} \right);\left( { - 2; - 3} \right)} \right\}
\end{array}$
