Since $$x^2-xy+2y^2-116=0$$ are defined over the real integers, we know that the discriminant of the equation (if we solve it for x) will be greater than or equal to zero.
Thus,
$$(-y)^2-4(1)(2y^2-116) \ge 0$$
$$464-7y^2 \ge 0$$
$$(\sqrt{464}+\sqrt{7}y)(\sqrt{464}-\sqrt{7}y) \ge 0$$ $$\rightarrow {-8.1416 \le y \le 8.1416} $$
But y will be an integer, thus, we know that $${-8 \le y \le 8} $$ must be true.
From the given equation $$x^2-xy+2y^2=116$$
We know that we can manipulate the RHS of the equation by doing the following:
$$x^2-xy+2y^2=114+2(1)$$ which implies $$y=\pm1$$
This gives $$x^2-x(\pm 1)=114$$ but this leads to non-integer solutions for x.
We repeat the process from $$y=2$$ to $$y=8$$ (we will cover the negative values of y as well because of the fact that $$y^2=1, 4, 9, 14, 25, 36, 49, 64$$ would cover the y values from $$y=\pm1, \pm2, \pm3, \pm4, \pm5, \pm6, \pm7, \pm8$$) and we find that all of the integer solutions to the equations are $$(6, -5), (-11, -5), (11, 5), (-6, 5), (2, -7), (-9, -7), (9, 7), (-2, 7), (-2, -8), (-6, -8), (6, 8), $$ and $$(2, 8)$$ which gives us the range of $$xy$$ as $$-30 \le xy \le 63$$.
(Edit:I now notice that this solution is similar to that posted by Albert, which I did not see until after making my post...I am sorry!)