First, the problem has no solution in that it is unbounded below. Let $x_\epsilon = -3 -\epsilon, y_\epsilon = 7+\epsilon$, with $\epsilon>0$. Then $(x_\epsilon, y_\epsilon)$ satisfies the constraint and the cost goes to $-\infty$ as $\epsilon \to 0$.
However, finding stationary points of the Lagrangian is pretty straightforward, yielding $\pmatrix{-\frac{1}{(3+x_1)^2} \\ - \frac{7}{(2+x_2)^2}}+ \lambda \pmatrix{1 \\ 1 } = 0$ This gives $\frac{1}{(3+x_1)^2} = \lambda = \frac{7}{(2+x_2)^2}$, from which we obtain $|3+x_1| = \frac{1}{\sqrt{7}}|2+x_2|$. There are only two possibilities to consider, (1) $3+x_1 = \frac{1}{\sqrt{7}}(2+x_2)$ and (2) $3+x_1 = -\frac{1}{\sqrt{7}}(2+x_2)$.
Solving these results in (1) $x = \frac{-9+3 \sqrt{7}}{2}, y = \frac{17-3 \sqrt{7}}{2}$, and (2) $x = \frac{-9-3 \sqrt{7}}{2}, y = \frac{17+3 \sqrt{7}}{2}$. Substituting these values in gives (1) $\frac{2\sqrt{7}}{4\sqrt{7}-7}$ and (2) $\frac{2\sqrt{7}}{4\sqrt{7}+7}$. However, only (1) could qualify as a minimum in the sense that it does minimize the cost on the set $(-3,\infty)\times (-2,\infty)$ (subject to the constraint, of course).