How to find the non-differentiable point(s) of a given continuous function?

If we define f[x] e.g. like this:

f[x_] := Abs[x]

the following returns interesting points:

Reduce[ 
    Limit[(f[x + h] - f[x])/h, h -> 0, Assumptions -> x ∈ Reals, Direction -> -1] != 
    Limit[(f[x + h] - f[x])/h, h -> 0, Assumptions -> x ∈ Reals, Direction -> 1],  x]

x == 0

Let's try another function defined with Piecewise, e.g.

g[x_] := Piecewise[{{x^2, x < 0}, {0, x == 0}, {x, 1 > x > 0}, 
                    {1, 2 >= x >= 1}, {Cos[x - 2] + x - 2, x > 2}}]

then we needn't use Assumptions in Limit:

Reduce[ Limit[ (g[x + h] - g[x])/h, h -> 0, Direction -> -1] != 
        Limit[ (g[x + h] - g[x])/h, h -> 0, Direction -> 1], x]

 x == 0 || x == 1 || x == 2

pts = {x, g[x]} /. {ToRules[%]};

Plot[ g[x], {x, -5/4, 3}, PlotStyle -> Thick, 
      Epilog -> {Red, PointSize[0.023], Point[pts]}]

One should be careful when working with Piecewise since Reduce may produce errors when weak inequalities (LessEqual) are involved. For this reason we added {0, x == 0} in the definition of the function g.

Here is an approach that you can use for numerical functions that at least have a left and right derivative. If such a function isn't differentiable in a point that is equivalent to the left and right derivatives being unequal, so look at the left and right finite difference approximation of the derivative, and see where they disagree.

rightd[f_, h_, x_] := (f[x + h] - f[x])/h
leftd[f_, h_, x_] := (f[x] - f[x - h])/h

f[x_?NumericQ] := If[x < .13, 0 , x - .13 ]

Plot[Abs[leftd[f, 0.1, x] - rightd[f, 0.1, x]], {x,-1,1}, PlotRange->All]

Using this you can use some numerical maximization on Abs[leftd[f, h, x] - rightd[f, h, x]], perhaps with successively smaller h to avoid false positives.