how could inverse be found of this $x^ne^{x^2}$

Question

$f:\mathbb{R}\to [0,\infty), f(x)=x^n e^{x^2}, n\in \mathbb{N}$ I need to find $(f^{-1})'(e)$, also to show $f^{-1}$ is continously differentiable,

First of all I tried to find inverse like this: if $g$ be the inverse then, $f(g(x))\equiv I_{[0,\infty)}=x\Rightarrow (g)^ne^{g^2}=x $ but I think if I take log both side , I can not find explicictly $g$ , pplease help.

Answer 1

Hint:

use $$ \left(f^{-1} \right)'(x)=\frac{1}{f'(f^{-1}(x))} $$

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with $x=e$. And note that $f(1)=e$.