It's a problem found with the help of WA .
Let $0<x$ a real number and $n\geq 1$ a natural number then we have : $$ f(1)\leq f(x)=(1+x)^{\frac{-1}{(nx)}}+\Big(1+\frac{1}{x}\Big)^{-\frac{x}{n}}<f(0)$$
I have also conjectured that :
Let $1\leq x$ a real number and $n\geq 1$ a natural number then we have : $$f\Big(\frac{x+\frac{1}{x}}{2}\Big)\leq f(x)$$
This conjecture I think is useful because of this fact :
Let $g(x)=\frac{x+\frac{1}{x}}{2}$ and $x\geq 1$ a real number then we have :
$$f(1) \leq f(g_n(x))\leq f(g_{n-1}(x))\leq \cdots \leq f(g(x))\leq f(x)\leq f(g^{-1}(x))\leq \cdots\leq f(g^{-1}_n(x))<f(0)$$
Where we speak about the inverse of the function $g(x)$ and the iteration ($n$ to $n$ times) of the function $g(x)$ with itself .
So the idea is to prove more generaly that $x=1$ is a minimum and $x=0$ is an infimum .
Well I have tried the same method as here by user Robin Aldabanx (first answer with bounty) for the first case or $n=1$ .
I have tried power series as well without success this time .
I have been also inspired by the Polya's proof of Am-Gm but no good issues .
Update
Case $n=1$
One can prove that the function :
$$h(x)=(1+x)^{\frac{-1}{x}}$$
is concave on $(0,\infty)$ . So we can apply Karamata's inequality and a majorization to get something of the kind :
$$h(x)+h\Big(\frac{1}{x}\Big)\geq h(y)+h\Big(\frac{1}{y}\Big)\quad (1)$$
The inequality $(1)$ gives information on $f(x)$ in the case where $n=1$.Via the majorization we can say as it's increasing or decreasing .
Moreover I think we can apply this method to the general case $n\geq 1$.
I don't know if it's really relevant but the inverse function of $h(x)$ is :
$$h^{-1}(x)=\frac{\operatorname{W}(x\log(x))-\log(x)}{\log(x)}$$
With the Lambert's function .
If you have a nice way to solve it .
Thanks you very much .