i haven't followed the notebook "closely". but if mathematica can't Integrate or NIntegrade f[x,y] to show the (volume), then how will you (use mathematica) to determine if any ellipse is within %95 of the correct answer which was not yet found?

In[1113]:= Clear[x, y, t]

In[1114]:= x[t_] := t;

In[1115]:= y[t_] := t;

In[1116]:= Integrate[
 1 + 1.5 10^-3 x[t] + 2.5 10^-3 x[t]^2 + 1.7 10^-3 y[t] + 
  1.8 10^-3 y[t]^2 + 1.8 10^-3 x[t] y[t], t]

Out[1116]= t + 0.0016 t^2 + 0.00203333 t^3

or on the other hand,

In[1117]:= Integrate[
 1 + 1.5 10^-3 x + 2.5 10^-3 x^2 + 1.7 10^-3 y + 1.8 10^-3 y^2 + 
  1.8 10^-3 x y, {x, 0, 1}, {y, 0, 1}]

Out[1117]= 1.00348

what i'm not seeing here is why you don't use a double integral (for volume) which is solvable, or set up an implicit equation and use implicit rules

can either or both of x and y be independant? can they both be written that way? are functions allows as parameters to this function or only numbers? it's possible for some functions of f[x,y] to be written in the form of a family of ellipses (with multple as argument). are you intending that f[x,y] be any function, or is f[x,y] the only function you need evaluated?

i think a little more needs to be know about your specifications before one can conclude regions

Notice first that the problem can be simplified, since function[0, 0]=1, and one can compute K=NIntegrate[LKLH[x, y], {x, -100, 100}, {y, -100, 100}]. Now, we can define, with the same function:

Truc[nth_] := nth Exp[-nth];
LKLH[x_, y_] := Truc[function[x, y]]/K;

Since "function" is the equation of a conic, we can study it.

Pol = Expand@FullSimplify@(function[x, y] - L)
M = CoefficientList[%, {x, y}];
règles = CoefficientRules[Pol, {x, y}];

Put it in canonical form:

{a, b, c, d, e, 
   f} = {{2, 0}, {1, 1}, {0, 2}, {1, 0}, {0, 1}, {0, 0}} /. règles;
b^2 - 4 a c

Aq = {{a, b/2, d/2}, {b/2, c, e/2}, {d/2, d/2, f}};
MatrixForm@Aq
 Det[Aq] // Factor

Since the determinant is negative for L>= 1, its is the equation of an ellipse under this condition. Suppose now L>1; The solution of Pol==0 is the contour : function[x,y]==L. I think you can now write formally:

\[Integral]Truc[
    function[x, y]] \[DifferentialD]x \[DifferentialD]y = \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(+\[Infinity]\)]\(Truc[
    L] \[DifferentialD]L\)\)

Then, if it's true, you just have to compute:

Q = Quantile[GammaDistribution[2, 1], 0.95]

since Truc is indeed this Gamma distribution, Then, determine the ellipse, and plot it.

Ellips = Solve[Expand@FullSimplify@(function[x, y] - Q) == 0, {x, y}];
Print@Show[{ParametricPlot[{x, y /. Ellips[[1]]}, {x, -50, 50}, 
     PlotStyle -> Red], 
    ParametricPlot[{x, y /. Ellips[[2]]}, {x, -50, 50}, 
     PlotStyle -> Green]}, PlotRange -> All];
Glob = ContourPlot[function[x, y], {x, -70, 50}, {y, -70, 70}, 
  PlotPoints -> 50, Contours -> {1.003, 1.05, 1.5, 2, 5.5}, 
  ContourShading -> None, ContourLabels -> True, PlotRange -> All]
Print@Show[
   ContourPlot[function[x, y], {x, -70, 70}, {y, -70, 70}, 
    Contours -> {Q}, ContourShading -> None, ContourLabels -> True, 
    ContourStyle -> Red, PlotRange -> All]
   , Glob, PlotRange -> All];

I think the region of interest is the interior of the red ellipse (if there is no mistake in my reasoning!).

try evaluating it with numerical arguments

try testing your LKLH function.