In my opinion $f_n(x) = \ln \left(e^x + \frac{1}{n} \right)$ is not uniform convergent. Of course we have $f_n \rightarrow \ln e^x = x$. Then we have to find $$d_{sup}(f_n,f) = \sup_{x \in [0,1]} |f_n(x) - f(x) = \sup_{x \in [0,1]} \left( \ln \left(e^x + \frac{1}{n} \right) - \ln \left(e^x \right) \right)$$
If we take $$h(x) = \ln \left(e^x + \frac{1}{n} \right) - \ln \left(e^x \right) $$ then we have $$h'(x) = \frac{-1}{ne^x + 1}$$ which is not equal $0$ for all $x \in [0,1]$.
So we should search supremum in $x=0$ or $x=1$
$$h(1) = \ln \left( e + \frac{1}{n} \right) > h(0)$$ so $$\sup_{x \in [0,1]} \left( \ln \left(e^x + \frac{1}{n} \right) - \ln \left(e^x \right) \right) = \ln \left( e + \frac{1}{n} \right) \rightarrow 1 \not=0$$. Hence $f_n$ is not uniform convergent.
Could you tell me, am I right? Does it work?
You found $h'(x)<0$ so the function $h$ is decreasing so $$||f_n-f||_\infty=h(0)=\ln\left(1+\frac 1 n\right)\to 0$$ so the convergence is uniform on $[0,1]$.