The function $y = ||x| - 1|$ is differentiable for all values of '$x$' except

The given function is $y = ||x| - 1|$.

A function of the form $|f(x)|$ is generally non-differentiable at points where $f(x) = 0$ (if the curve crosses the x-axis) and at points where $f(x)$ itself is non-differentiable.

First, the inner function $f(x) = |x| - 1$ involves $|x|$, which is non-differentiable at $x = 0$.
For $x$ in the neighborhood of $0$ (i.e., $-1 < x < 1$), $y = ||x| - 1| = 1 - |x|$.
The left-hand derivative at $x = 0$ is $1$ and the right-hand derivative is $-1$. Thus, $y$ is non-differentiable at $x = 0$.

Second, we check where the expression inside the outer modulus becomes zero:
$|x| - 1 = 0$
$\Rightarrow |x| = 1$
$\Rightarrow x = 1$ or $x = -1$

At $x = 1$, for $x > 1$, $y = x - 1$ (derivative is $1$). For $0 < x < 1$, $y = 1 - x$ (derivative is $-1$). Thus, $y$ is non-differentiable at $x = 1$.

At $x = -1$, for $x < -1$, $y = -x - 1$ (derivative is $-1$). For $-1 < x < 0$, $y = x + 1$ (derivative is $1$). Thus, $y$ is non-differentiable at $x = -1$.

Therefore, the function is differentiable for all values of $x$ except at $x \in \{-1, 0, 1\}$.

Answer: $\{-1 , 0 , 1\}$