Curious result related to the function $f(x)=\exp\Big(\frac{x-1}{x}\ln(3)\Big)$ Let me defines somethings :


Let $0<x<1$ let $f(x)$ be the function :
    $$f(x)=\exp\Big(\frac{x-1}{x}\ln(3)\Big)$$
    And :
    $$g(x)=f(1-x)$$
    Denote by :
    $$\min_{x\in(0,1)}(f(x)+g(x))=\frac{2}{3}$$
    Define $f^n(x)$ by ($n\geq2$ a natural number):
    $$f^n(x)=\underbrace{f(f(f(f(\cdots(x)\cdots)}_{n \quad \text{times}}$$
    And $g^n(x)$ by :
    $$g^n(x)=g(\underbrace{g(1-g(1-g(1-g(\cdots g(1-x))\cdots)}_{(n-1) \quad \text{times}}$$
    Then we have
    $$\min_{x\in(0,1)}(f^n(x)+g^n(x))=\frac{2}{3}$$


Let me show 
$$\min_{x\in(0,1)}(f(x)+g(x))=\frac{2}{3}$$
The derivative is :
$$f'(x)+g'(x)= 3^{\frac{(x - 1)}{x}} \Big(\frac{1}{x} - \frac{(x - 1)}{x^2}\Big) \log(3) + 3^{\frac{-x}{1-x}} \Big(-\frac{x}{(1 - x)^2} - \frac{1}{(1 - x)}\Big) \log(3)$$
Or :
$$f'(x)+g'(x)=\frac{\Big(3^{\frac{(x - 1)}{x}} (x - 1)^2 - 3^{\frac{-x}{1-x}} x^2\Big) \log(3)}{((x - 1)^2 x^2)}$$
Now it's not hard to show that the derivative vanishes at $x=0.5$ and get the desired result using the closed interval method . 
For the other cases I believe that there is a trick or somethings like that .
Thanks a lot for your time and patience .
Ps:I think furthermore that we can replace the value $\ln(3)$ be something more general $\ln(\alpha)$ by example with $\alpha\geq 2$  . 
 A: In the following I will use a subscript notation $f_n, g_n$ instead of the given superscript one as it might lead to confusion regarding the exponentiation.
At first one needs to note that the direct definitions of $f_n, g_n$ can be rewritten in terms of recursive definitions.
For $f_n$ it is rather obvious $f_n(x) = f(f_{n-1}(x))$ whereas for $g_n$ it might help to write down the first couple of terms
\begin{align*}
  g_2(x) &:= g(g(1-x)) = g(f(x))
  \\
  &= g(f_1(x))
  \\
  g_3(x) &:= g(g(1-g(1-x)))= g(g(1-f(x)))= g(f(f(x)))
  \\
  &= g(f_2(x))
  \\
  g_4(x) &:= g(g(1-g(1-g(1-x))))
  \\
  &= g(f_3(x))
\end{align*}
which gives the general recursive formula
$$
  g_n(x) = g(f_{n-1}(x)) \,.
$$
(To be exact I definied $f_1 := f$.)
Next we need to differentiate $f_n+g_n$ to find its extrema
\begin{align*}
  (f_n(x)+g_n(x))'
  &= f_n'(x)+g_n'(x)
  \\
  &= (f(f_{n-1}(x)))' + (g(f_{n-1}(x)))'
  \\
  &= f'(f_{n-1}(x)) \cdot f_{n-1}'(x) + g'(f_{n-1}(x)) \cdot f_{n-1}'(x)
  \\
  &= f_{n-1}'(x) \left[ f'(f_{n-1}(x)) + g'(f_{n-1}(x)) \right]
  \\
  &= f_{n-1}'(x) \left[ f'(y) + g'(y) \right] \,, \qquad y := f_{n-1}(x)
  \\
  &\overset{!}{=} 0
\end{align*}
where I introduced the definition of $y$ to make it more obvious that both terms in the sum depend on the same variable.
As $f_{n-1}'(x) \neq 0$ we need to solve
$$
  0 = f'(y) + g'(y)
$$
and the solution $y=y_0=\frac{1}{2}$ has been shown in the first part.
Compute the extremal value ($x_0$ refers to $y_0=f_{n-1}(x_0)$)
\begin{align*}
  f_n(x_0)+g_n(x_0)
  &= f(y_0)+g(y_0)
  = \frac{2}{3}
\end{align*}
where the summation value follows from the first part.
For the argument to be complete note that $f$ is one-to-one.
The same goes for $f_n$.
Is it still a minima (we only showed that its derivative is zero at $x_0$)?
Note that $f_n+g_n$ is continuous and the limit for $x\rightarrow0^+$ is easily computed to be 1.
Thus, it it still a minima.
A: We have 
$\forall x\in (0,1), f(x)+g(x)=3^{-\frac{1-x}{x}}+3^{-\frac{x}{1-x}}$
The function $\varphi:x\mapsto \frac{1-x}{x}$ is a bijection from (0,1) to $(0,+\infty)$
Moreover, 
$\forall x\in (0,1), f(x)+g(x)=3^{-\varphi(x)}+3^{-\frac{1}{\varphi(x)}}$
We consider the function $h:x\mapsto 3^{-x}+3^{-\frac{1}{x}}$ on $(0,+\infty)$
The derivative is given by: $h'(x)=\frac{3^{-1/x}\ln(3)-3^{-x}\ln(3)x^2}{x^2}$
$\forall x\in(0,+\infty), h'(x)>0\Leftrightarrow 3^{-1/x}>3^{-x}x^2$
$\Leftrightarrow -\frac{1}{x}\ln(3)>-x\ln(3)+2\ln(x)$
$\Leftrightarrow \frac{1}{x}\ln(3)-x\ln(3)+2\ln(x)<0$
Study the function $\psi:x\mapsto 2\ln(x)-x\ln(3)+\frac{\ln(3)}{x}$
The dérivative $\psi'(x)=-\frac{x^2\ln(3)-2x+\ln(3)}{x^2}$ is negative on $(0,+\infty)$
So $\psi$ is strictly decreasing and $\psi(1)=0$
$\forall x\in (0,+\infty), h'(x)>0\Leftrightarrow \psi(x)<0\Leftrightarrow x>1$
$\forall x\in (0,+\infty), h'(x)=0\Leftrightarrow \psi(x)=0\Leftrightarrow x=1$
Thus h is strictly decreasing on (0,1) and striclty increasing on $(1,+\infty)$. Moreover h is continuous, so $h$ admits a minimum at 1.
Hence $f+g$ admits a minimum at $\varphi^{-1}(1)=\frac12$.
For the second part I resume the work of @jack
We have
$\forall x\in(0,1), f^n(x)+g_n(x)=f(f^{n-1}(x))+g(f^{n-1}(x))$
We remark that $f((0,1))\subseteq (0,1)$ so $f^{n-1}((0,1))\subseteq (0,1)$ 
We deduce that
$\forall x\in(0,1), f(f^{n-1}(x))+g(f^{n-1}(x))\geq \min_{x\in(0,1)} (f(x)+g(x))$
Then
$\forall x\in(0,1), f^n(x)+g_n(x)\geq \frac{2}{3}$
Thus $\inf_{x\in(0,1)} (f^n(x)+g_n(x))\geq \frac{2}{3}$
However the function $f$ is a bijection from (0,1) to (0,1)
By considering $a=f^{-1}\left(f^{-1}\left(...f^{-1}\left(\frac12\right)...\right)\right)$ (n-1 times), we have $a\in(0,1)$ and
$f^n(a)+g_n(a)=f(f^{n-1}(a))+g(f^{n-1}(a))=f\left(\frac12\right)+g\left(\frac12\right)=\frac32$
Hence the infimum is reached and it is a minimum
$\min_{x\in(0,1)} (f^n(x)+g_n(x))=\frac{2}{3}$
