How to do this: ∫∫[-∞->∞] e^(-x^2-xy-y^2) dx dy?

This is a good exercise in change of variables (ie "substitution") in multiple integrals.

First make a change of variable x = u + v, y = u - v to kill the xy term:

x^2 + xy + y^2 = (u+v)^2 + (u+v)(u-v) + (u-v)^2

= u^2 + 2uv + v^2 + u^2 - v^2 + u^2 - 2uv + v^2

= 3 u^2 + v^2

The Jacobian associated to this change of variables is constant, in fact J(u,v) = 2 for all u and v. So your integral is equal to

2 * the integral over R^2 of e^(-(3u^2 + v^2)) du dv.

Now introduce a new variable s by u = (1/sqrt(3)) s (leave v alone). The advantage of doing this is that u^2 = (s^2)/3 so that 3u^2 = s^2 and hence -(3u^2 + v^2) = -(s^2 + v^2). The Jacobian of this change of variable is 1/sqrt(3), so your integral is now

(2/sqrt(3)) * the integral over R^2 of e^(-(s^2 + v^2)) ds dv

This integral you presumably already know (s and v are dummy variables; you can rename them x and y if you like and maybe it will look more familiar); it's pi. If you don't know it, you can do it by a change of variables to polar coordinates: let s = r cos t and y = r sin t, the Jacobian of this transformation is r, so the integrand becomes e^(-r^2) * r, and the region of integration is now the strip [0,oo] x [0, 2pi] in the (r,t) plane. So it's

(2/sqrt(3)) * the integral from 0 to 2pi [the integral from 0 to infinity of r e^(-r^2) dt] dt

A simple substitution shows that the inner integral in [ ] is 1/2, so the integral from 0 to 2pi of that is just pi, so the whole thing is just (2/sqrt(3)) * pi.