Differentiability of f everywhere but |f| nowhere, f: R - > R?

I ran across an interesting problem in Spivak's Calculus on Manifolds
- I am unaware of whether or not it is solvable.  Maybe someone
perusing this board will know the answer.

The problem:

Exhibit a differentiable function f:R->R such that the function |f|
defined by

|f|(t) = |f(t)|

is not differentiable.  (Spivak, Calculus on Manifolds 2-13d)

It is a little bit mystifying to me exactly what he is looking for
here.  Unless the problem is requiring that the derivative exists
nowhere on R it seems uniquely unchallenging (i.e. very unlike a
Spivak problem, for the most part).

So I was wondering if anyone knew of such a function that is as
described above, with the assumption that |f|'(t) exists nowhere on R.

Full context of the problem follows, and it seems to me that they
should be related to this.  Just assume each of parts (a),(b), and (c)
to be true (or solve them if you wish but I am only concerned with the
above - part (d)).

2-13 Define IP:R^n X R^n -> R by IP(x,y) = <x,y>

a)  Find D(IP)(a,b) and (IP)'(a,b) [Note: here Spivak is making a
distinction between the linear transformation D(IP)(a,b) and it's
matrix with respect to the usual basis of R^n, which he generally just
calls f'(a), in this case(IP)'(a,b)]

b)  If f,g:R -> R^n are differentiable and h:R->R is defined by 
h(t) = <f(t),g(t)>, show that h'(a) = <f'(a)^T,g(a)> + <f(a),g'(a)^T>
(here the T indicates the transpose).

c)  If f:R -> R^n is differentiable and |f(t)| = 1 for all t, show
that
<f'(t)^T,f(t)> = 0