I'm trying to find the convergence of $f_n$ and $f_n'$ where $f_n(x)=\frac{1}{1+nx^2}$.
From the function if I derivate the result is $f'n= -\frac{2nx}{(1+nx^2)^2}$. To determine $f$ I have to take the limit, in order to have $f(x)=\lim_{n\to\infty}f_n(x)$, in the first case like $f_n$ converges to $0$; while for $f_n'$ the limit seems to be 0 as well (taking the limit respect $n$ and having a $n^4$ in the denominator).
Now with the uniform convergence: $f$ is zero then is uniformly convergent I have to see if $\lim_n \sup |\frac{1}{1+nx^2}|=0$. $f'n=0 \iff x=0$, but is not zero a saddle point? Then the conclusion is that $f_n$ is convergent but is not uniformly convergent?
You are almost correct in saying $\displaystyle\lim_{n\to\infty}f_n(x) = 0$, but you should consider what happens for $x = 0$. To determine uniform convergence, note that a sequence of continuous functions which converge uniformly, must converge to a continuous function. Based on what I said above, Is $\displaystyle\lim_{n\to\infty}f_n(x)$ continuous? If not, you can conclude that the convergence is non-uniform.