\[{\rm{Let\, }}f\left( x \right) = \left\{ \begin{array}{l} {\left( {x - 1} \right)^2}\cos \frac{1}{{x - 1}} - \left| x \right|,\,x \ne 1\\ - 1,\,x = 1 \end{array} \right.\]
The set of points where $$f\left( x \right)$$  is not differentiable is :

Solution :
The doubtful points are $$x=0,\,1$$   because these are the turning points of the definition. $$\left| x \right|$$ is not differentiable at $$x=0,$$   while $${\left( {x - 1} \right)^2}\cos \frac{1}{{x - 1}}$$    is differentiable. The algebraic sum of a differentiable function and a nondifferentiable function is nondifferentiable. So, $$f\left( x \right)$$  is not differentiable at $$x=0.$$  Now,
$$\eqalign{ & f'\left( {1 + 0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h - 1} \right)}^2}\cos \frac{1}{{1 + h - 1}} - \left| {1 + h} \right| - \left( { - 1} \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}\cos \frac{1}{h} - h}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \left( {h\cos \frac{1}{h} - 1} \right) \cr & = 0 - 1 = - 1 \cr} $$
Similarly, $$f'\left( {1 - 0} \right) = - 1.$$    So, $$f\left( x \right)$$  is differentiable at $$x=1.$$