Showing that the sequence of functions is not Cauchy I need to show that $ g_n(x)=x^{1/(2n-1)}  $ is not a Cauchy sequence in $C[-1,1] $ w.r.t. supremum norm. I tried to find the maximum of the difference of $g_n$ and $g_m$ by just differentiating but that did not really lead me anywhere.

Any suggestions would be greatly appreciated.
If you look at the graphs of $x^{1/n}$ for some large values of $n$, you should see a clear pattern in their behavior on $[0, 1]$: it looks as though they converge pointwise to:

$$f(x)=\begin{cases} 0\ \text{ if $x=0$} \\ 1\ \text{ else} \end{cases}$$

Of course with odd $n$ we get symmetrical behavior on $[-1, 0]$. Note that this convergence is pointwise, and not with respect to the supremum norm. Now, as pointed out in the comments, if you can show that $g_n$ does not converge, you will show that it is not Cauchy, so this suggests a plan of attack. If you remember the following fact, you can use what I've said to prove that $g_n$ does not converge w.r.t. the supremum norm:


  If $f_n$ converges to $f$ w.r.t. the supremum norm, then $f_n$ converges pointwise to $f$.
If you look at the graphs of $x^{1/n}$ for some large values of $n$, you should see a clear pattern in their behavior on $[0, 1]$: it looks as though they converge pointwise to:

$$f(x)=\begin{cases} 0\ \text{ if $x=0$} \\ 1\ \text{ else} \end{cases}$$

Of course with odd $n$ we get symmetrical behavior on $[-1, 0]$. Note that this convergence is pointwise, and not with respect to the supremum norm. Now, as pointed out in the comments, if you can show that $g_n$ does not converge, you will show that it is not Cauchy, so this suggests a plan of attack. If you remember the following fact, you can use what I've said to prove that $g_n$ does not converge w.r.t. the supremum norm:


  If $f_n$ converges to $f$ w.r.t. the supremum norm, then $f_n$ converges pointwise to $f$.