Derivative of the absolute value $\vert x\vert^3$ with $\epsilon$-$\delta$-definition - check my proof Consider $$f: \mathbb{R} \to \mathbb{R}, ~~ f(x) = \vert x \vert ^3$$ Show that the first derivative exists and it holds $$f'(x)=3x\vert x\vert ~~\forall x \in \mathbb{R}.$$ 
My approach:
edited
For a $y > 0$: I want to show that $3y^2$ is the limit of $\frac{f(x)-f(y)}{x-y}$. This means for any given $\epsilon >0$ I find a $\delta >0$ such that we have $ \forall x \in \mathbb{R}_{>0}~:\vert x - y\vert < \delta \Rightarrow \biggl|\frac{x ^3 - y^3}{x-y}- 3y^2\biggr| < \epsilon$. After some manipulations and a polynomial long division I get the following: $\biggl|\frac{x ^3 - y^3}{x-y}- 3y^2\biggr| =~ ... ~= \vert x^2-2y^2+xy\vert$. But now I am stuck. How do I use this to show that the conition is satisfied?
I run into the same problem when I try the other case:
For $y<0$: I want to show that $-3y^2$ is the limit of $\frac{f(x)-f(y)}{x-y}$. This means for any given $\epsilon >0$ I find a $\delta >0$ such that we have $ \forall x \in \mathbb{R}_{<0}~:\vert x - y\vert < \delta \Rightarrow \biggl|\frac{y ^3 - x^3}{x-y}+ 3y^2\biggr| < \epsilon$. After some manipulations and a polynomial long division I get the same as in the "$y>0$-case": $\biggl|\frac{y ^3 - x^3}{x-y}+ 3y^2\biggr|=\biggl|\frac{x ^3 - y^3}{x-y}- 3y^2\biggr| =~ ... ~= \vert x^2-2y^2+xy\vert$. How do I use this to show that the conition is satisfied?
However, the last case seems to be the easiest one...
If $y=0$ then I have to take a closer look at the limit. I will use the $\epsilon$ -$\delta$-definition of the limit. So if a value $b$ is the limit then for any given $\epsilon >0$ I find a $\delta >0$ such that we have $ \forall x \in \mathbb{R}~:\vert x - 0\vert < \delta \Rightarrow \biggl|\frac{\vert x \vert ^3 - 0}{x-0}-b \biggr| < \epsilon$. I will now set $b=0$ and check if the condition is satisfied.
$ \biggl|\frac{\vert x \vert ^3 - 0}{x-0}-0 \biggr| =  x^2$. Setting $\delta= \sqrt{\epsilon}$ shows that the limit exists and hence $f$ is differentiable at $0$.
Did I use the $\epsilon$ -$\delta$-definition of the limit rigorously?

No, since you only used the $\varepsilon-\delta$ definition at $0$ (correctly, by the way, although I would have taken $\delta=\sqrt\varepsilon$). In all other cases, what you did was to “apply the rules of differentiation”.
