For an alternate proof, note that $f_n'(x)=0$ implies $x=\pm\frac1n$, and \begin{align} f_n\left(\frac1n\right) &= \frac1{2n}\\\ f_n\left(-\frac1n\right) &= -\frac1{2n}\\\ f(-1) &= -\frac1{1+n^2}\\\ f(1) &= \frac1{1+n^2}. \end{align} It follows that $$\lim_{n\to\infty}\sup_{x\in[-1,1]}|f_n(x)|=0, $$ so that $f_n$ converges uniformly to $0$ on $[-1,1]$.