Problem :
Let the function on $x\in(0,1),a=\operatorname{constant}>1$ :
$$f(x)=x^{a^{x^{a}}}a^{x^{a^{x}}}-\operatorname{arctanh}(x)$$
Now let :
$$f'(u)=0,0.99<u<1$$
Then it seems we have :
$$\lim_{a \to +\infty} (a-f(u)-\ln(a))=C$$
Where $C$ is a constant
Question :
Does this constant have a closed form or a nice approximation ?
The first estimate of $u$, solution of $f'(u)=0$, is obtained by the simplest series expansion to $O(x-1)$ $$u_0=1-\frac{2}{4 a^2 (1+\log (a))-1}$$ which, by Darboux theorem, is an over estimate of the solution.
The next estimate $u_1$ will be explicitly given by the first iteration of any Newton like method of order $n$.
For example, for $a=10$
$$\left( \begin{array}{cc} u_0 & 0.998484887553 \\ u_1^{(2)} & 0.998352785241 \\ u_1^{(3)} & 0.998339082166 \\ u_1^{(4)} & 0.998338974554 \\ \cdots & \cdots \\ \text{solution} & 0.998338974249 \\ \end{array} \right)$$
However, successive Newton iterations are quite better
$$\left( \begin{array}{cc} u_0 & 0.998484887553 \\ u_1^{(2)} & 0.998352785241 \\ u_2^{(2)} & 0.998339099648 \\ \cdots & \cdots \\ \text{solution} & 0.998338974249 \\ \end{array} \right)$$
I shall use as solution $u=u_2^{(2)}$.
Computing $$F(a)=a-f(u)-\log(a)$$ a lot of overflows happen as soon as $a >26$.
Nevertheless, Mathematica generates nice plots and the function $a\,F(a)$ looks to be a straight line with a slope close to $2.5$.
However, this is far to be sure since, using $u=u_0$, we have $$\color{blue}{F(a)=\frac{1+2 \log (2)}{2} +\frac{1}{2} \log (1+\log (a))-}$$ $$\color{blue}{\frac{\log ^2(a)+5 \log (a)+1}{8 a (1+\log (a))^2}+O\left(\frac{1}{a^2}\right)}$$ which is bery good even for small values of $a$.
Notice that $$\int \log (1+\log (a))\,da=a \log (1+\log (a))-\frac 1 e \text{Ei}(1+\log (a))$$ making, for $10^3 \leq a \leq 10^6$, the average value of the integrand equal to $2.62353$ and $$\frac{1+2 \log (2)}{2} +\frac{1}{2}\times 2.62353=2.50491$$
Probably $$\lim_{a \to +\infty} (a-f(u)-\ln(a))$$ does not exist.
Edit
Let $(u_2,u_3,u_4,\cdots)$ be the first iterate of Newton, Halley and Householder,$\cdots$ methods (obtained starting at $u=0$). Asymptotically, we have
$$u_{n+1}-u_n=-\frac{ 1}{2^{n}\, (n-1)!\,\,a^{n+1}\, \log (a) }$$
Let $$G_k(a)=a-f(u_k)-\log(a)$$
Expanded as series to $O\left(\frac{1}{a^2}\right)$ for large values of $a$, they all lead to the "blue" formula.
Using more terms and $L=\log(a)$ $$G_4(a)=\frac{1+2 \log (2)}{2} +\frac{1}{2} \log (1+L)-\frac{L^2+5 L+1}{8 (L+1)^2\,a}-$$ $$\frac{14 L^4+38 L^3+62 L^2+13 L+2}{48 (L+1)^4\,a^2}-$$ $$\frac{57 L^6+520 L^5+1122 L^4+1479 L^3+352 L^2+81 L+9}{384 (L+1)^6\,a^3}+O\left(\frac{1}{a^4}\right)$$
Asymptotically $$G_2(a)-G_3(a) \sim \frac{1}{16 a^2}$$
If there is an asymptotic, it would be $$\underset{a\to \infty }{\text{limit}}\Bigg(\frac{a-f(u)-\log (a)}{\log (1+\log (a))}\Bigg)=\frac 12$$