The function \( f(x) = ||x| - 1| \) is analyzed through sequential transformations. The initial function \( y = |x| \) exhibits a vertex at \( x = 0 \). \[\text{The graph of } y = |x| \text{ has a sharp corner at } x = 0.\] Shifting down by one unit, \( y = |x| - 1 \) results in a vertical translation. \[\text{The graph shifts vertically downward by 1 unit.}\] Applying the absolute value to the entire expression, \( f(x) = ||x| - 1| \), generates sharp corners at specific points. \[\text{Sharp corners are formed at } x = \{-1, 0, 1\}.\] Consequently, \( f(x) \) is not differentiable at \( \{-1, 0, 1\} \). Final Answer: \[\boxed{\pm 1, 0}\]