is $f\left(x\right)\:=\:\left|x\right|^3$ twice differentiable? Consider the function $f\left(x\right)\:=\:\left|x\right|^3$ , $f:\mathbb{R}\rightarrow \mathbb{R}$.
1) Is it twice differentiable? And if so, how can I prove this and calculate it?
2) If it does, can I conclude it has third differentiable for every $x$?  
So far, for 1, I believe is true but I really stuck in the proof:
$$\lim _{h\to 0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)\:=\:\lim _{h\to 0}\left(\frac{\left|x+h\right|^3-\left|x\right|^3}{h}\right)\:=\:\lim_{h\to 0}\left(\frac{\left(\left|x+h\right|^{\:}-\left|x\right|\right)\left(\left|x+h\right|^2\cdot \left|x+h\right|\left|x\right|+\left|x\right|^2\right)}{h}\right)\:$$ But how to continue from here? I also tried this approach:  $$\lim _{x\to x_0}\left(\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}\right)\:=\:\lim_{x\to x_0}\left(\frac{\left|x\right|^3-\left|x_0\right|^3}{x-x_0}\right)\:=\:\lim_{x\to x_0}\left(\frac{\left(\left|x\right|^{\:}-\left|x_0\right|\right)\left(\left|x\right|^2+\left|x\right|\left|x_0\right|+\left|x_0\right|^2\right)}{x-x_0}\right)\:\:$$
I have serious problem with the next step. Can someone guide me what tricks I need to do? Thanks in advance!

You can show that $f'(x) = 3x|x|$ by looking at the cases $x > 0$ (where $f(x) = x^3$), $x < 0$ (where $f(x) = -x^3$) and $x = 0$ (use the definition of the derivative). 
Then show that $f''(x) = 6|x|$ by looking at the cases $x > 0$ (where $f'(x) = 3x^2$), $x < 0$ (where $f'(x) = -3x^2$) and $x = 0$ (use the definition of the derivative again).  
Thus $f''(x)$ exists everywhere. Try this one more time and you will find that $f'''(0)$ does not exist. Hence $f'''$  does not exist everywhere. 
