I am searching for the functions $f,g:[0,\infty) \to [0,\infty)$ which are both increasing, $f'$ is strictly increasing, $g$ is the inverse of $f$ and
$$ g(x)^2=f(x)^2\cdot g'(x)$$
My approach was to get rid of $f$ by plugging $x=g(x)$ and obtain something which contains only $g$:
$$ (g\circ g)^2(x) = x^2 \cdot g' \circ g(x)$$
but this seems to lead nowhere. Do you have some ideas?
I can't even find an example, since $f(x)=x$ doesn't work because $f'$ should be strictly increasing...
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \totald{}{x}\bracks{1 \over {\rm g}\pars{x}} = -\,{1 \over \fermi^{2}\pars{x}} \quad\imp\quad {1 \over {\rm g}\pars{x}} = -\int{\dd x \over \fermi^{2}\pars{x}} + \mbox{a constant}. $$