Looking for differentiable functions $f$ such that the set of points at which $|f|$ is not differentiable has some particular properties I know that $\sin x$ is differentiable at all points but $|\sin x|$ is not differentiable at countably many points namely at integer multiples of $\pi$ . So I am asking the following questions 
i) Give example of a differentiable function $f:\mathbb R \to \mathbb R$ such that the set of points at which $|f|$ is  not differentiable is countable and dense in $\mathbb R$ 
ii) Give example of a differentiable function $f:\mathbb R \to \mathbb R$ such that the set of points at which $|f|$ is  not differentiable is uncountable and not dense in $\mathbb R$ 
iii) Give example of a differentiable function $f:\mathbb R \to \mathbb R$ such that the set of points at which $|f|$ is  not differentiable is uncountable and  dense in $\mathbb R$ 
$ UPDATE$:- As mentioned in the comments if $f$ is differentiable and and at some point non-zero then since by continuity $f$ will have same sign in a  neighborhood of that point , $|f|$ will be differentiable ; thus $|f|$ is not differentiable $c$ only when $f(c)=0$ but then as  John pointed out , if such points are dense then this leads to a contradiction , resolving i) and iii) in the negative ; this leaves us with (ii) only  
 A: I CLAIM THAT THE ANSWER TO (ii) IS NEGATIVE. PROOF.
Lemma: If $f$ is differentiable over $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ and $c$ is a SIMPLE zero of $f$ then it is isolated, it means there
exists $\varepsilon >0$ such that $for$ all $x\in \left( c-\varepsilon
,c+\varepsilon \right) ,$ $f(x)=0$ iff $x=c.$ Proof. Assume the contrary
holds. Then there exists a sequence $x_{n}$ of zeros of $f$ converging to $c,
$ then $$f^{\prime }(c)=\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}%
=\lim_{n\rightarrow \infty }\frac{f(x_{n})-f(c)}{x_{n}-c}=\lim_{n\rightarrow
\infty }\frac{0-0}{x_{n}-c}=0.$$ Then $$f(c)=f^{\prime }(c)=0.$$ Contradicting
the simplicity of the zero $c.$
It follows that the Answer to (ii) is negative. In fact, every isolated subset
of $%
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%BeginExpansion
\mathbb{R}
%EndExpansion
$ is countable.
A: This answer is incomplete but gives some results and extends the observation that the answer to questions (i) and (iii) is negative.
If $f:\mathbb R\to\mathbb R$ is differentiable, $|f|$ fails to be differentiable at a point $x\in\mathbb R$ if and only if $f(x)=0$ and $f'(x)\neq0$.
Continuity of $f$ follows from differentiability.
We are interested in the set $A=f^{-1}(0)\setminus(f')^{-1}(0)$ where $|f|'$ doesn't exist.
Some results:


*

*$A$ is nowhere dense (its closure has empty interior).
Reason:
Suppose this is not the case.
Then there is an open interval $I$ so that $A\cap I$ is dense in $I$.
In particular this implies that $f^{-1}(0)\cap I$ is dense in $I$, which by continuity implies that $f|_I\equiv0$.
It follows that also $f'|_I\equiv0$, a contradiction.

*$A$ need not be closed even if $f'$ is continuous.
Example:
$f(x)=x^3\cos(1/x)$ for $x\neq0$ and $f(0)=0$.
Now $f$ is continuously differentiable and $0\in\partial A\setminus A$.
Zeros of $f$ accumulate at zero but the derivative doesn't vanish at other zeros.

*If $f'$ is continuous, $A$ has zero measure.
Reason:
If $f$ is continuously differentiable, it is locally Lipschitz, and so is $|f|$.
Lipschitz functions are differentiable almost everywhere by Rademacher's theorem, so the set of nondifferentiability has to have zero measure.


It seems that $A$ should be something like a Cantor set if we want it to be uncountable.
I faintly recall seeing a paper in arXiv where the author(s?) constructs a function which vanishes on the ternary Cantor set but its derivative is nonzero in this set.
Unfortunately I couldn't find the paper, and advice against taking my memory as a decisive argument.
