The relationship between the derivative of $f(x)$ and $|f(x)|$ I have seen it in an exercise book. I don't know how to do it.

If $f(x)$ is continuous at $x=a,$ and $|f(x)|$ can be differentiated at $x=a,$ then $f(x)$ is differentiable at $x=a.$

 A: Let $g(x)=|f(x)|$. If $f(a)\ne 0$, then by the continuity of $f$, there is an interval $I_\epsilon=(a-\epsilon, a+\epsilon)$ such that if $x$ is in $I_\epsilon$, then  $f(x)$ has the same sign as $f(a)$. If $f(a)$ is positive, then for all $x$ in $I_\epsilon$, we have $g(x)=f(x)$. So for $x$ in the interval $I_\epsilon$,
$$\frac{g(x)-g(a)}{x-a}=\frac{f(x)-f(a)}{x-a}.$$
It follows that if $\lim_{x\to a}\frac{g(x)-g(a)}{x-a}$ exists, then so does $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ (and the limits are the same). We conclude that $f'(a)$ exists and is equal to $g'(a)$.
If $f(a)$ is negative, then for all $x$ in $I_\epsilon$, we have $g(x)=-f(x)$. It follows that
$$\frac{g(x)-g(a)}{x-a}=\frac{-f(x)+f(a)}{x-a}=-\frac{f(x)-f(a)}{x-a}.$$
Thus if  $\lim_{x\to a}\frac{g(x)-g(a)}{x-a}$ exists, then so does $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$. In this case, $f'(a)=-g'(a)$.
Now we come to the interesting case, where $f(a)=0$.
We are told that $\lim_{x\to 0}\frac{g(x)-g(a)}{x-a}$ exists. Since $g(a)=0$, this says that $\lim_{x\to a}\frac{g(x)}{x-a}$ exists.  
But $g(x)\ge 0$ for all $x$. So for all $x>a$, we have $\frac{g(x)}{x-a}\ge 0$. It follows that 
$$\lim_{x\to a+}\frac{g(x)}{x-a} \ge 0.\tag{$1$}$$
Also, if $x&lta$, then $x-a&lt0$, and therefore $\frac{g(x)}{x-a}\le 0$. It follows that
$$\lim_{x\to a-}\frac{g(x)}{x-a} \le 0.\tag{$2$}$$
Since by assumption $\lim_{x\to a}\frac{g(x)}{x-a}$ exists, 
we conclude from inequalities $(1)$ and $(2)$ that
$$\lim_{x\to a}\frac{g(x)}{x-a} = 0.\tag{$3$}$$
It follows immediately that 
$$\lim_{x\to a}\frac{f(x)}{x-a} = 0.$$
This says that $f'(a)$ exists and is equal to $0$.
