Answer
False
Work Step by Step
The statement is false.
Consider $f(x)=x$ and $a=0$.
$\lim\limits_{x \to 0^-}\frac{|f(x)|-|f(0)|}{x-0}=\lim\limits_{x \to 0^-}\frac{|x|-|0|}{x-0}=\lim\limits_{x \to 0^-}\frac{-x-0}{x}=-1$
but
$\lim\limits_{x \to 0^+}\frac{|f(x)|-|f(0)|}{x-0}=\lim\limits_{x \to 0^+}\frac{|x|-|0|}{x-0}=\lim\limits_{x \to 0^+}\frac{x-0}{x}=1$
$f$ is differentiable at $0$ but $|f(x)|=|x|$ is not differentiable at $0$ because $\lim\limits_{x \to 0^-}\frac{|f(x)|-|f(0)|}{x-0}\neq \lim\limits_{x \to 0^+}\frac{|f(x)|-|f(0)|}{x-0}$.
