I've been sitting here for hours trying to find a solution to his problem:
If you have the function $g(y)$, which is the inverse of $$f(x) = x^x\\ e^{-1} \leq x < \infty,$$ Show that $$\lim\limits_{y \to \infty} \dfrac{g(y) \ln (\ln y)}{\ln y} = 1$$
Hint: Start with the equation $y=x^x$ and take the ln of both sides twice.
If we let that $y = x^x$ then $g(y) = x$ because that $g(x^x) = x$ by the definition of the problem. Then if we use the hint given by the problem
$$\ln(y) = x\ln(x)$$
So, for $e^{-1} \leq x < \infty \implies 0< e^{-x} \leq x^x < \infty$ we have that $0 < y < \infty$ so take $x > 1$ then
$$\lim_{y \to +\infty}\frac{g(y)\ln(\ln(y))}{\ln(y)} = \lim_{x \to +\infty}\frac{x\ln(x\ln(x))}{x\ln(x)} = \lim_{x \to +\infty}\frac{\ln(x\ln(x))}{\ln(x)} = \lim_{x \to +\infty}\left(1 + \frac{1}{\ln(x)}\right) = \color{red}{1}$$
Because $\ln$ is a crescent - and not bounded above - function we used that
$$\lim_{x \to +\infty}\frac{1}{\ln(x)} = 0$$