<s>[INST] Functions satsify $f'=f^{-1}$ with $f^{-1}$ is compostional inverse of$ f$ let $f$ be a function such that :$f:\mathbb{C}\to \mathbb{C}$ and $f^{-1}$ is the compositional inverse of $f$, I seek for the analyticity of $f$ at $0$, then my question here is :


Question:
    Are there functions satisfy:$f'=f^{-1}$ with  $f^{-1}$  is compostional inverse of  $f$ ?


 [/INST] If
$f'=f^{-1}$
and $f(x) = ax^b$,
then
$f'(x) = ab x^{b-1}$
and
$f^{(-1)}(x)
=(x/a)^{1/b}
$
so
$(x/a)^{1/b}
=ab x^{b-1}
$.
If $b \ne 0$,
then,
raising to the $b$ power,
$x/a
=(ab)^b x^{b(b-1)}
$
or
$a^{-b-1}b^{-b}
= x^{b(b-1)-1}
$.
For the right side
to be constant,
we must have
$0
=b(b-1)-1
=b^2-b-1
$
so
$b
=\dfrac{1\pm \sqrt{1+4}}{2}
=\dfrac{1\pm \sqrt{5}}{2}
$
and
$a^{-b-1}b^{-b}
= 1
$
or
$a^{-b-1}
= b^b
$
or,
since
$b+1 = b^2$,
$a
=b^{\frac{b}{-b-1}}
=b^{\frac{b}{-b^2}}
=b^{\frac{-1}{b}}
$.
</s>