2. All real number pairs $(x, y)$ that satisfy the equation
$$
(x+3)^{2}+y^{2}+(x-y)^{2}=3
$$

are $\qquad$ .
2. $(-2,-1)$.

Expanding and rearranging the given equation yields $x^{2}+3 x+y^{2}-x y+3=0$. Completing the square gives $\left(y-\frac{x}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2} x+\sqrt{3}\right)^{2}=0$.
From this, we get $x=-2, y=-1$.