This can be done with help of Mathematica as follows. Let us exhaust the plane by disks centered at the origin. Switching to the polar coordinates, one obtains
Integrate[r*Exp[a*x^2 + b*x*y + c*y^2] /. {x -> r*Cos[ϕ],y -> r*Sin[ϕ]}, {r, 0, R}]
(-1 + E^(R^2 (a Cos[ϕ]^2 + b Cos[ϕ] Sin[ϕ] + c Sin[ϕ]^2)))/(a + c + (a - c) Cos[2 ϕ] + b Sin[2 ϕ])
It is clear the finite limit of the function defined by the above expression as R->Infinity exists only if ForAll[ϕ, ϕ > -Pi && ϕ <= Pi, a Cos[ϕ]^2 + b Cos[ϕ] Sin[ϕ] + c Sin[ϕ]^2 < 0] is valid as the result of
Integrate[r*Exp[a*x^2 + b*x*y + c*y^2] /. {x -> r*Cos[ϕ], y -> r*Sin[ϕ]}, {r, 0,Infinity},Assumptions -> {a, b, c} ∈ Reals && ϕ > -Pi && ϕ <= Pi]
ConditionalExpression[-(1/( 2 (a Cos[ϕ]^2 + b Cos[ϕ] Sin[ϕ] + c Sin[ϕ]^2))), a Cos[ϕ]^2 + b Cos[ϕ] Sin[ϕ] + c Sin[ϕ]^2 < 0]
demonstrates.
Unfortunately, Mathematica fails with
Resolve[ForAll[ϕ, ϕ > -Pi && ϕ <= Pi, a Cos[ϕ]^2 + b Cos[ϕ] Sin[ϕ] + c Sin[ϕ]^2 < 0], Reals]
, returning the input. However, Mathematica is able to derive the condition in such a way
Resolve[ForAll[y, y >= -1 && y <= 1,a y^2 + b *y*Sqrt[1 - y^2] + c (1 - y^2) < 0], Reals]
c < 0 && a < b^2/(4 c)
Now
Integrate[(-1)/(a + c + (a - c) Cos[2 ϕ] + b Sin[2 ϕ]), {ϕ, -Pi, Pi}, Assumptions -> c < 0 && a < b^2/(4 c) && a < 0 && b ∈ Reals]
results in
ConditionalExpression[ -Piecewise[{{0, (Sqrt[Abs[-(a/(a - I*b - c)) - c/(a - I*b - c) - Sqrt[-b^2 + 4*a*c]/(a - I*b - c)]] < 1 && Sqrt[Abs[-(a/(a - I*b - c)) - c/(a - I*b - c) + Sqrt[-b^2 + 4*a*c]/(a - I*b - c)]] < 1) || (Sqrt[Abs[-(a/(a - I*b - c)) - c/(a - I*b - c) - Sqrt[-b^2 + 4*a*c]/(a - I*b - c)]] >= 1 && Sqrt[Abs[-(a/(a - I*b - c)) - c/(a - I*b - c) + Sqrt[-b^2 + 4*a*c]/(a - I*b - c)]] >= 1)}, {(2*Pi)/Sqrt[-b^2 + 4*a*c], Sqrt[Abs[-(a/(a - I*b - c)) - c/(a - I*b - c) + Sqrt[-b^2 + 4*a*c]/(a - I*b - c)]] < 1}}, (-2*Pi)/Sqrt[-b^2 + 4*a*c]], !(Sqrt[Abs[-(a/(a - I*b - c)) - c/(a - I*b - c) - Sqrt[-b^2 + 4*a*c]/(a - I*b - c)]] >= 1 && Sqrt[Abs[-(a/(a - I*b - c)) - c/(a - I*b - c) - Sqrt[-b^2 + 4*a*c]/(a - I*b - c)]] <= 1) && !(Sqrt[Abs[-(a/(a - I*b - c)) - c/(a - I*b - c) + Sqrt[-b^2 + 4*a*c]/(a - I*b - c)]] >= 1 && Sqrt[Abs[-(a/(a - I*b - c)) - c/(a - I*b - c) + Sqrt[-b^2 + 4*a*c]/(a - I*b - c)]] <= 1)]
, thinking a few hours. In fact, this is $$\frac{2 \pi }{\sqrt{4 a c-b^2}} $$ as Simplify under the conditions c < 0 && a < b^2/(4 c) && a < 0 && b ∈ Reals] shows.
Integrate[Exp[a*x^2 + b*x*y + c*y^2], {x, -d, d}, {y, -d, d}, Assumptions -> {a, b, c} \[Element] Reals && d > 0, GenerateConditions -> False]returns a one-dimensional integral. $\endgroup$Integrate[Exp[a*x^2 + b*x*y + c*y^2], {x, -d, d}, {y, -d, d}, Assumptions -> {a, b, c} > 0 && d > 0]. $\endgroup$