Let us first study the pointwise convergence of the sequence $ f_n $. For this we will consider the following limit, $$\lim_{n\to \infty}n\ln \left( \dfrac{1+nx}{nx} \right )=\lim_{n\to \infty}\dfrac{\ln \left (\dfrac{1+nx}{nx} \right )}{\dfrac{1}{n}}. $$ Applying the L'Hospital rule $$\lim_{n\to \infty}\dfrac{\ln \left (\dfrac{1+nx}{nx} \right )}{\dfrac{1}{n}}=\lim_{n\to \infty}\dfrac{n}{nx+1}=\dfrac{1}{x+1/n}=\dfrac{1}{x}.$$ Therefore $f_n\to \dfrac{1}{x}$ pointwise when $n\to \infty$ at $(0,\infty)$.
Now, for each $n\in \Bbb N$ and $x>0$, define $A_n$ \begin{align*} A_n&:=\sup_{x\in (0,\infty)}\{|f(x)-f_n(x)|\}\\ &=\sup_{x\in (0,\infty)}\left \{\dfrac{\left |1-nx\ln\left (\dfrac{1+nx}{nx} \right)\right |}{|x|} \right \}\\ &\ge \sup_{x\in (0,\infty)}\left \{\dfrac{|1-nx\ln(1+nx)}{|x|} \right \}\\ &\ge \sup_{x\in (0,\infty)}\left \{\dfrac{|1-nx\ln(2)}{|x|} \right \}\\ &\ge \sup_{x\in (0,\infty)}\left \{\dfrac{|1-x\ln(2)|}{|x|} \right \}\\ &=\sup_{x\in (0,\infty)}\left \{\left |\dfrac{1}{x}-\ln(2) \right | \right \}\to \ln(2),x\to\infty. \end{align*}
So $ A_n \ge \ln (2) $. Thus, $ f_n $ does not converge uniformly. Is my try right ?