Function satisfying $f(x)^{f^{-1}(x)}=x^2$ with $ f^{-1}$ is a compositional inverse of $f$ and $f:\mathbb{R+}\to \mathbb{R+}$?

Question

Let $f$ be a function such that :$f:\mathbb{R+}\to \mathbb{R+}$ and $f^{-1}$ is a compositional inverse of $f$ , I have tried to find solution of the following functional $f(x)^{f^{-1}(x)}=x^2$, I took $f(x)=x$ but it doesn't work it coincide only for $x=1$ and for $f(x)=\exp(x)$ I come up to $x^x=x^2$ then coincide only for $x=1$ and $x=2$ , so $f(x)^{f^{-1}(x)} > x$ for $x >2$ which means no trivial solution exists probably a formel power series exist arround $x=1$ or $x=2$,Then my question here is : How I can solve $f(x)^{f^{-1}(x)}=x^2$ with $ f^{-1}$ is a compositional inverse of $f$ ?

Edit I suspect such trigonometric function would be works for the titled function,like $\tan x$ as shown here but this need a restriction of our Domain of definitions for which $\tan x$ to be defined

Edit I have edited the question according to the answer given by @Robert I without changing the meaning I missed that I search about a real valued function not complex as stated in the below answer

Answer 1

When $f(x) < 0$ (which must happen for uncountably many $x$ if $f^{-1}$ is defined on $\mathbb R$), the fact that $f(x)^{f^{-1}(x)} > 0$ requires $f^{-1}(x)$ to be a rational number (presuming we define $a^b = \exp(b \log(a))$ for some branch of the logarithm; note all branches of logarithm of a negative number have imaginary part an integer multiple of $\pi i$, and $\exp(b k \pi i)$ is a positive real only when $b k$ is an even integer). Since $f$ can't map a countable set onto an uncountable one, we conclude there is no such function.

Or did you want to just define $f$ and $f^{-1}$ on $(0,\infty)$?