Problem: Let $E$ be a normed space, and $f:[a,b]\to E$ be continuous. Define $g:[a,b]\to\mathbb R$ by $g(x)=\|f(x)\|$. Prove that if $f'_{+}(t_0)$ exists for some $t\in[a,b)$, then so does $g'_{+}(t_0)$, and moreover $$|g'_{+}(t_0)|\leq \|f'_{+}(t_0)\|.$$
What I have done: I couldn't prove the existence of $g'_{+}(x)$, however assume it exists I can prove the inequality as follows: For $h>0$, we have \begin{align*} g(t_0+h)-g(t_0) &=\|f(t_0+h)-f(t_0)+f(t_0)\|-\|f(t_0)\|\\ &\leq \|f(t_0+h)-f(t_0)\| \end{align*} Since $\|\cdot\|$ is continuous, and by assumption $g'_{+}(t_0)$ exists, we obtain that $g'_+(t_0)\leq \|f'_+(t_0)\|.$ Similar argument yields $-g'_+(t_0)\leq \|f'_+(t_0)\|$. Thus we obtain the desired inequality.
Need Help: How to prove $g'_+(t_0)$ exist?
Suppose $x \in \mathbb{R}, e_1,e_2 \in E$, then the map $\phi:[a,b] \mapsto \mathbb{R}$ given by $\phi(x) = \|e_1+xe_2\| $ is convex and hence has a right derivative at any point in $[a,b)$ (see here).
Let $\gamma = f_+'(x)$, $r(h) = f(x+h)-f(x)-h\gamma$ and suppose $h>0$ (and $x+h \le b$).
Note that $\|f(x+h)\| = \|f(x)+h\gamma + r(h) \| $ satisfies
$\|f(x)+h\gamma \| + \| r(h) \| \ge \|f(x+h)\| $ and $\|f(x+h)\| \ge \|f(x)+h\gamma\| - \|r(h) \| $.
By subtracting $\|f(x)\|$ and dividing both sides by $h$ and letting $h \downarrow 0$ we get $\lim_{h \downarrow 0} {\|f(x)+h\gamma \| - \|f(x)\| \over h} = \lim_{h \downarrow 0} {\|f(x+h) \| - \|f(x)\| \over h}$ (recall that $\phi$ defined above has a right derivative). In particular, $g_+'(x)$ exists.