A function $$f$$ is defined on $$[-3, 3]$$ as
$$f(x) = \begin{cases} \min\{|x|, 2 - x^2\}, & -2 \leq x \leq 2 \\ [|x|], & 2 < |x| \leq 3 \end{cases}$$
where $$[x]$$ denotes the greatest integer $$\leq x$$. The number of points, where $$f$$ is not differentiable in $$(-3, 3)$$ is ______.

For $$-2 \leq x \leq 2$$, we have $$f(x) = \min\{|x|, 2 - x^2\}$$. Setting $$|x| = 2 - x^2$$ gives $$x^2 + |x| - 2 = 0$$, so $$|x| = 1$$, meaning $$x = \pm 1$$. For $$|x| < 1$$: at $$x = 0$$, $$|x| = 0 < 2 = 2 - x^2$$, so $$f(x) = |x|$$. For $$1 < |x| < 2$$: at $$x = 1.5$$, $$|x| = 1.5 > 2 - 2.25 = -0.25$$, so $$f(x) = 2 - x^2$$.

For $$2 < |x| \leq 3$$, $$f(x) = [|x|]$$ (greatest integer function). So for $$2 < x \leq 3$$, $$f(x) = [x] = 2$$, and for $$-3 \leq x < -2$$, $$f(x) = [|x|] = 2$$.

Now we identify points of non-differentiability in $$(-3, 3)$$:

At $$x = 0$$: $$f(x) = |x|$$ has a corner (left derivative $$= -1$$, right derivative $$= 1$$), so $$f$$ is not differentiable.

At $$x = 1$$: for $$x < 1$$, $$f(x) = x$$ with derivative 1. For $$x > 1$$, $$f(x) = 2 - x^2$$ with derivative $$-2x = -2$$. Since $$1 \neq -2$$, not differentiable. Similarly at $$x = -1$$: left derivative of $$2-x^2$$ is $$-2(-1) = 2$$, right derivative of $$|x| = -x$$ is $$-1$$. Not differentiable.

At $$x = 2$$: $$f(2) = \min\{2, -2\} = -2$$ from the left, but $$f(x) = [x] = 2$$ for $$x$$ just above 2. Since $$f(2^-) = -2 \neq 2 = f(2^+)$$, there is a discontinuity, hence not differentiable.

At $$x = -2$$: similarly $$f(-2) = \min\{2, -2\} = -2$$ and $$f(x) = 2$$ for $$x$$ just below $$-2$$. Discontinuity, hence not differentiable.

No other points of non-differentiability exist (the greatest integer function $$[|x|] = 2$$ is constant on $$(2, 3)$$ and $$(-3, -2)$$, so it is differentiable there).

The total number of points of non-differentiability is $$5$$.