Let S be the set of points where the function $$f(x) = |2 - |x - 3||$$, $$x \in R$$, is not differentiable. Then $$\sum_{x \in S} f(f(x))$$ is equal to

We are given the function $$f(x)=\left|\,2-\bigl|x-3\bigr|\,\right|$$ for all real numbers $$x$$. A point where a function contains the symbol $$|\,\cdot\,|$$ can fail to be differentiable in only one of two situations:

1. The outer absolute value changes from the form $$|u|=u$$ to $$|u|=-u$$, that is, when its argument $$u$$ becomes $$0$$.
2. The inner expression itself is an absolute-value function such as $$|x-3|$$, which is not differentiable at the point where its own argument is zero.

We therefore inspect the two absolute values present.

(i) Zero of the inner absolute value.
The inner symbol is $$|x-3|$$, which becomes zero when $$x-3=0$$, i.e. at $$x=3$$. Because $$|x-3|$$ is not differentiable at that point, the composite function $$f(x)$$ can also fail to be differentiable at $$x=3$$.

(ii) Zero of the outer absolute value.
The argument of the outer absolute value is $$2-|x-3|$$. Setting it equal to zero, we get $$2-|x-3|=0\;\Longrightarrow\;|x-3|=2 \;\Longrightarrow\;x-3=\pm2 \;\Longrightarrow\;x=1\;\text{ or }\;x=5.$$ Thus $$x=1$$ and $$x=5$$ are two more candidate points of non-differentiability.

No other values of $$x$$ make either absolute value switch slope, so the set of points where $$f(x)$$ is not differentiable is $$S=\{1,\,3,\,5\}.$$

We now need the sum $$\sum_{x\in S} f\!\bigl(f(x)\bigr) =f\!\bigl(f(1)\bigr)+f\!\bigl(f(3)\bigr)+f\!\bigl(f(5)\bigr).$$ To compute these terms smoothly, we first write $$f(x)$$ without nested absolute values. Let us consider the value of $$|x-3|$$ and then of $$f(x)=|2-|x-3||$$ step by step.

When $$|x-3|\le 2$$ (that is, $$1\le x\le 5$$) the inner quantity $$2-|x-3|$$ is non-negative, so the outer absolute value does nothing and $$f(x)=2-|x-3|.$$ When $$|x-3|>2$$ (that is, $$x<1$$ or $$x>5$$) the inner quantity is negative, so the outer absolute value adds a minus sign and $$f(x)=|x-3|-2.$$ Next we remove the remaining absolute value $$|x-3|$$ to obtain four linear pieces:

For $$1\le x\le 3$$: here $$|x-3|=3-x,$$ so $$f(x)=2-(3-x)=x-1.$$ For $$3\le x\le 5$$: here $$|x-3|=x-3,$$ so $$f(x)=2-(x-3)=5-x.$$ For $$x>5$$: here $$|x-3|=x-3,$$ so $$f(x)=x-3-2=x-5.$$ For $$x<1$$: here $$|x-3|=3-x,$$ so $$f(x)=3-x-2=1-x.$$

Using these explicit formulas we evaluate $$f(x)$$ at the three points in $$S$$.

At $$x=1$$ (lies in the interval $$1\le x\le 3$$):
$$f(1)=1-1=0.$$ At $$x=3$$ (lies in either $$1\le x\le 3$$ or $$3\le x\le 5$$, both give the same value):
$$f(3)=3-1=2 \quad\text{or}\quad f(3)=5-3=2.$$ At $$x=5$$ (lies in $$3\le x\le 5$$):
$$f(5)=5-5=0.$$

Now we must feed these results back into the function once more.

For $$x=1$$ we need $$f\!\bigl(f(1)\bigr)=f(0).$$
Since $$0<1,$$ the appropriate formula is $$f(x)=1-x,$$ giving $$f(0)=1-0=1.$$ For $$x=3$$ we need $$f\!\bigl(f(3)\bigr)=f(2).$$
Because $$2\in[1,3],$$ we again use $$f(x)=x-1,$$ so $$f(2)=2-1=1.$$ For $$x=5$$ we need $$f\!\bigl(f(5)\bigr)=f(0),$$ already evaluated above as $$1.$$

Adding the three results, we obtain $$\sum_{x\in S} f\!\bigl(f(x)\bigr)=1+1+1=3.$$ So, the answer is $$3$$.