Exercise of differentiable functions in $\mathcal{C}[0,1]$ Consider $E=\mathcal{C}[0,1]$ with norm $\|\cdot\|_\infty$. For which $x$ is differentiable the following functions:
a)  $f:E\rightarrow E$ defined by $f(x)(t)=|x(t)|^{2/3}$
b)  $f:E\rightarrow \mathbb{R}$ defined by $f(x)=||x||_{\infty}$
 A: 1) For  $x$ that satisfy $x(t)\ne 0$ for all $t$. Same proof as in your earlier question Differentiable function in the normed space $\,\mathcal{C}[0,1]$ which had $1/2$ in place of $2/3$. I hope you are not going to ask about $3/4$ next.   
2) For  $x$ such that the set $A=\{t\in [0,1]:|x(t)|=\|x\|_\infty\}$ consists of one point. The idea of the proof given by Hagen von Eitzen to your earlier equation Differentiability of the supremum norm in $\ell^{\infty}$ carries over to the continuous case, although more effort is required here. 
Indeed, suppose $A$ contains at least two points $t_1,t_2$. Let $z$ be a continuous function such that $z(t_1)=x(t_1)$ and $z(t_2)=-x(t_2)$. Then for all real $h$
$$f(x+hz) \ge \max_{i=1,2} |x(t_i)+hz(t_i)| \ge (1+|h|)f(x)$$ 
Since $f(x+hz)-f(x)\ge |h|f(x)$, it follows that the limit  $\lim_{h\to 0}(f(x+hz)-f(x))/h$ does not exist.
If $A=\{t_0\}$, then $x(t_0)\ne 0$. The derivative $D_xf$ is the linear operator $D_xf(z)=z(t_0)\operatorname{sign} x(t_0)$, which I leave for you to prove. 
