I have the sequence $$f_n(x)=\log \left(\Pi_{k=1}^n \left(\frac{x^k+1}{x^k}\right)^k\right)$$ and I have to study the uniform convergence.
Equivalently we have $$f_n(x)=\sum_{k=1}^n \frac{k}{x^k} \;\log \left(1+\frac{1}{x^k}\right)^{x^k}$$
Now the original request tranforms in studying the uniform convergence of the series $$\sum_k \frac{k}{x^k} \;\log \left(1+\frac{1}{x^k}\right)^{x^k}$$
The first thing I want to do is verifying the necessary condition: necessary condition for the series to converge is that $$\lim_k \frac{k}{x^k} \;\log \left(1+\frac{1}{x^k}\right)^{x^k}=0$$
If we take any $x$ in $(-\infty,-1) \cup(1,+\infty)$ the above equality holds. (If I have done the limit correctly).
Now I want to try with the ratio test: $$\lim_k \; \left|\frac{(k+1)\log\left(1+\frac{1}{x^{k+1}}\right)}{k\log\left(1+\frac{1}{x^k}\right)}\right|=\frac{1}{|x|}$$ which assures convergence if $x\in (-\infty,-1) \cup(1,+\infty)$. (If I have done the limit correctly).
For the uniform convergence I have to study $$\sup \left| \sum_{k=1}^{+\infty} k\log\left(1+\frac{1}{x^k}\right)-\sum_{k=1}^{n} k\log\left(1+\frac{1}{x^k}\right)\right|=\sup \left| \sum_{k=n+1}^{+\infty} k\log\left(1+\frac{1}{x^k}\right)\right|$$ with $x\in I=(-\infty,-1) \cup(1,+\infty)$.
I noticed that this sup is greater than $(n+1)\log2$ and thus I have not uniform convergence in $I$.
Is my reasoning correct till here? How can I find the subset of $I$ in which I have uniform convergence?
When $x\in[-1,0]$ the sequence $f_n(x)$ is ill defined.
When $x\in(0,1]$, the general term of the series becomes unbounded as $k\to\infty$.
It follows that one need consider only $x\in I$. You've already shown pointwise convergence in $I$.
Now, for uniform convergence we need to show that given $\epsilon>0$ we can find $n\in\Bbb N$ such that for every $x\in I$
$$\left| \sum_{k=n+1}^{+\infty} k\log\left(1+\frac{1}{x^k}\right)\right| < \epsilon.$$
In other words, we can make the difference $|f_n(x) - f(x)|$ as small as desired by taking $n$ large enough uniformly on $x$.
For $z > -1$ we have
$$\log\left(1+z\right) \geqslant\frac{z}{1+z}$$
It follows that
\begin{align} \sup_{x\in I}\left|\sum_{k=n+1}^{+\infty} k\log\left(1+\frac{1}{x^k}\right)\right| &\geqslant \sup_{x\in (1,+\infty)}\left|\sum_{k=n+1}^{+\infty} k\log\left(1+\frac{1}{x^k}\right)\right| \\&= \sup_{x\in (1,+\infty)}\sum_{k=n+1}^{+\infty} k\log\left(1+\frac{1}{x^k}\right) \\&\geqslant \sup_{x\in (1,+\infty)}\sum_{k=n+1}^{+\infty} k\,\,\frac{x^{-k}}{1+x^{-k}} \\&= \sup_{x\in (1,+\infty)}\sum_{k=n+1}^{+\infty} \frac k {x^k + 1} \\&\geqslant \sup_{x\in (1,+\infty)}\sum_{k=n+1}^{+\infty} \frac k {2x^k} \\&= \sup_{x\in (1,+\infty)}\frac{(n+1)x-n}{2x^{n}{(x-1)}^2} \end{align}
As $x$ approaches $1$, we see that the sup is unbounded. Therefore, the $f_n$ fail to converge uniformly on $(1,+\infty)$.
On the other hand, for $z > -1$ we have
$$\log\left(1+z\right) \leqslant z.$$
Let $a>1$ and consider.
\begin{align} \sup_{x\in [a,+\infty)}\left|\sum_{k=n+1}^{+\infty} k\log\left(1+\frac{1}{x^k}\right)\right| &= \sup_{x\in [a,+\infty)}\sum_{k=n+1}^{+\infty} k\log\left(1+\frac{1}{x^k}\right) \\&\leqslant \sup_{x\in [a,+\infty)}\sum_{k=n+1}^{+\infty} \frac k {x^k} \\&= \sup_{x\in [a,+\infty)}\frac{(n+1)x-n}{x^n(x-1)^2} \end{align}
From the derivative of $g_n(x) = \frac{(n+1)x-n}{x^n(x-1)^2}$ being always negative and the fact that $\lim_{x\to\infty}g_n(x) = 0$, we conclude that
$$\sup_{x\in [a,+\infty)}\left|\sum_{k=n+1}^{+\infty} k\log\left(1+\frac{1}{x^k}\right)\right|\leqslant g_n(a).$$
Moreover, it's easy to see that $\lim_{n\to\infty}g_n(a) = 0$, so that the $f_n$ converge uniformly on $[a,+\infty)$.
The case for $x\in(-\infty,1)$ can be handled similarly, but requires some care dealing with absolute values. Do you think you can take it from here?