How to calculate or estimate the area of such implicit region about function $x^{\frac{1}{x+\frac{1}{x}}}$ How to estimate or calculate the area enclosed by the implicit equation.
$$x^{\frac{1}{x+\frac{1}{x}}}+y^{\frac{1}{y+\frac{1}{y}}}=\mathrm{e}$$
It is possible to prove that the area is between $13$ and $15$?
I draw the curve by Asymptote(http://asymptote.ualberta.ca/).
import graph;
import contour;
size(200);
real f(real x){return x^(1/(1/x + x));};
real g(real x, real y) {return f(x)+f(y);}
draw(contour(g,(.1,.1),(8,8),new real[]{exp(1)},operator ..),blue+2);
xaxis("$x$",BottomTop,LeftTicks);
yaxis("$y$",LeftRight,RightTicks);



 A: You have probably already solved it, but since it might be useful for someone else I write a couple of strategies using Wolfram Mathematica 13.0; obviously everyone can use any other spreadsheet.

Monte Carlo method
The basic idea is to bombard with 10^n balls the square of area $A_q$ which encloses the region of plane D of interest which goes from vertex (2,2) to vertex (7,7) and to count the m balls that also fall internally to D. The area of D is therefore estimated by $A_q\,m/10^n$. It goes without saying that a good pseudorandom real number generator is crucial here:
Table[m = 0; Do[{x, y} = RandomReal[{2, 7}, 2];
      If[x^(1/(x + 1/x)) + y^(1/(y + 1/y)) > E, ++m], {10^n}];
      25. m / 10^n, {n, 7}]


{12.5, 12.25, 13.675, 13.77, 13.7047, 13.7454, 13.7577}

or more efficiently:
Table[z = RandomReal[{2, 7}, {2, 10^n}];
      25. Total@Boole@Thread[Total[z^(1/(z + 1/z))] > E] / 10^n, {n, 8}]


{15., 15.5, 13.725, 13.87, 13.726, 13.7676, 13.7607, 13.7615}

The calculated area is the better the larger it's n.

Discrete Fourier Transform + Gauss-Green theorem
First, I tabulate the coordinates of some points of the boundary proceeding counterclockwise:
f[x_, y_] := x^(1/(x + 1/x)) + y^(1/(y + 1/y))

pts = Table[FindRoot[f[ρ Cos[θ] + 4, ρ Sin[θ] + 4] == E,
            {ρ, 1}][[1, 2]] {Cos[θ], Sin[θ]} + {4, 4},
            {θ, 0, 2 π, π/50}];

Through Discrete Fourier Transform, I determine a trigonometric fit of least squares to the data:
TrigFit[v_, m_, {x_, x0_, x1_}] := Module[{fc, fs, i, imax, j, n, t},
  n = Length[v]; imax = Min[m, Floor[(n - 1) / 2]]; t = 2 π (x - x0)/(x1 - x0);
  fc = Table[Sum[v[[j + 1]] Cos[2 π i j / n] / n, {j, 0, n - 1}], {i, 0, imax}];
  fs = Table[Sum[v[[j + 1]] Sin[2 π i j / n] / n, {j, 0, n - 1}], {i, 0, imax}];
  fc[[1]] + 2 Sum[fc[[i + 1]] Cos[i t] + fs[[i + 1]] Sin[i t], {i, 1, imax}]]

{x[t_], y[t_]} = TrigFit[pts, 50, {t, 0, 2 π}];

Finally, I calculate the area enclosed by this parametric curve using the Gauss-Green theorem:
NIntegrate[x[t] y'[t], {t, 0, 2 π}]


13.7592

If you want to check, it's also possible to superimpose the implicit curve on the parametric one:
Show[ParametricPlot[{x[t], y[t]}, {t, 0, 2 π}, PlotStyle -> Directive[Thick, Red]],
     ContourPlot[f[x, y] == E, {x, 0, 10}, {y, 0, 10}, PlotPoints -> 100]]


where it's evident that they overlap perfectly, so the calculated area is very close to the exact one!
