if neither f nor g is differentiable at x=a. is $f+g$ differentiable at $x=a$?
My answer: yes it is differentiable , because according to this question here: if neither f nor g is differentiable at x=a. is $f+g$ differentiable at $x=a$?
If $f$ is not differentiable then $-f$ is not differentiable (but I do not understand exactly why ? may be because f is not differentiable meaning that the limit of differentiability does not exist for f and hence it trivially does not exist also for -f ..... am I correct? ). now f+(-f) = 0 which is a differentiable function .... am I correct?
The condition of being non-differentiable is reflected in the inequality of Dini derivatives,
$$D^+f(a) = \limsup_{h \to 0+}\frac{f(a+h)-f(a)}{h} \\ D_+f(a) = \liminf_{h \to 0+}\frac{f(a+h)-f(a)}{h} \\ D^-f(a) = \limsup_{h \to 0-}\frac{f(a+h)-f(a)}{h}\\ D_-f(a) = \liminf_{h \to 0-}\frac{f(a+h)-f(a)}{h}$$
The four Dini derivatives always exist (possibly as extended real numbers), but only when all are equal is the function differentiable.
Note that $D^+f(a) = - D_+(-f)(a),$ etc., and it is then easy to show that if any two Dini derivatives of $f$ are not equal, then a pair of Dini derivatives of $-f$ are not equal.