jmc

By symmetry, the vertices of the square are $(\pm t,\pm t)$ for some positive real number $t.$ Then $$\frac{t^2}{3}+\frac{t^2}{6}=1.$$Solving, we find $t^2=2.$ Then $t=\sqrt{2}.$

The side length of the square is then $2t=2\sqrt{2},$ so its area is $(2\sqrt{2}{)}^2=\boxed{8}.$