$1.$ Suppose that $f:\mathbb{R} \rightarrow \mathbb{R}$ is diferentiable at $c$ and that $f(c)=0$. Show that $g(x):=|f(x)|$ is diferentiable at $c$ if and only if $f'(c)=0$.
$Proof.\Leftarrow $ Suppose that $f'(c)=0$ , then as $f$ is differentiable at $c$, we've got that: Let $\varepsilon>0$, there exist $\delta>0$ such that $x\in B_{\delta}(c)$ then:
$||\frac{f(c+h)}{h}|-|f'(c)||\leq |\frac{f(c+h)}{h}-f'(c)|<\varepsilon$, by hypothesis $f'(c)=0$, thus;
$\left | \frac{|f(c+h)|}{|h|} \right |<\varepsilon$, hence $g(x)$ is differentiable at $c$.
$\Rightarrow$ Suppose that $g(x):=|f(x)|$ is diferentiable at $c$, and suppose $f'(c)\not= 0$, then:
$g'(c)=lim_{x\to c}\frac{g(x)-g(c)}{x-c}=lim_{x \to c}\frac{|f(x)|-|f(c)|}{x-c}=lim_{h\to 0}\frac{|f(c+h)|}{|h|} $.
case 1. If $f'(c)>0$.
I have tried to reason for contradiction in this last implication, but I am not sure, I would like to receive a suggestion.
Assume $g'(c)$ exists. Then, since $f(c)=0$, the limit $$\lim_{h\to 0}\frac{|f(c+h)|}{h}$$ exists too. Call it $\alpha$. Clearly, the right-side limit is non-negative: $$\lim_{h\to 0,h\geq 0}\frac{|f(c+h)|}{h}\geq 0$$ so $\alpha\geq 0$. Clearly, the left-side limit is non-positive: $$\lim_{h\to 0,h\leq 0}\frac{|f(c+h)|}{h}\leq 0$$ hence $\alpha\leq 0$. So $\alpha$ must be zero. That is, $g'(c)=0$. This immediately implies that $f'(c)=0$.