$f(x) = \frac{e^{-|x|}}{max{(e^x,e^{-x})}} \\ \implies f(x) = \frac{e^{-|x|}}{e^{|x|}} \\ \implies f(x) = e^{-2|x|}$
Now we need to find the critical points
$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x} (e^{-2|x|}) \\ \implies f'(x) = \frac{-2xe^{-2|x|}}{|x|}$
Now we need to set $f'(x) = 0$ so
$\therefore \frac{-2xe^{-2|x|}}{|x|} = 0 \\ \implies -2xe^{-2|x|} = 0 \\ \implies e^{-2|x|} = 0$
$\mathrm{Critical\:points\:are\:points\:where\:the\:function\:is\:defined\:and\:its\:derivative\:is\:zero\:or\:undefined}$
x = 0 is the critical point
the left hand limit
$\lim_{x\rightarrow0^-}(f(x)) = \lim_{x\rightarrow0^-}(e^{-2|x|}) \\ = \lim_{x\rightarrow0^-}(e^{-2(-x)}) = \lim_{x\rightarrow0^-}(e^{2x}) = 1$
and the right hand limit
$\lim_{x\rightarrow0^+}(f(x)) = \lim_{x\rightarrow0^+}(e^{-2|x|}) \\ = \lim_{x\rightarrow0^+}(e^{-2(x)}) = 1$
we can see that both right hand and left hand limit is equal which means function is continuous.
For Differentiability we need to check for x = 0
$\lim_{x\rightarrow0^+}\frac{\partial }{\partial x}(e^{-2|x|}) = \lim_{x\rightarrow0^-}\frac{\partial }{\partial x}(e^{-2|x|}) \\ \implies \lim_{x\rightarrow0^+}\frac{-2xe^{-2|x|}}{|x|}= \lim_{x\rightarrow0^-}\frac{-2xe^{-2|x|}}{|x|} \\ \implies \lim_{x\rightarrow0^+}\frac{-2xe^{-2x}}{x}= \lim_{x\rightarrow0^-}\frac{-2xe^{-2(-x)}}{-x} \\ \implies -2 \neq 2$
$\therefore$ f(x) is not diffrentiable at x =0
C is the correct Answer