Function got a little lengthy for sake of symmetry!

We note that, $\dfrac{1}{c^s+c^{-e}+1}+\dfrac{1}{c^e+c^{-a}+1}=\dfrac{c^e}{c^{s+e}+c^e+1}+\dfrac{1}{c^e+c^{-a}+1}=\dfrac{c^e+1}{c^e+c^{-a}+1}\ \ \ (1).$ Because $s+e=-a$ . Next, since $-s=e+a)$ , $\dfrac{1}{c^a+c^{-s}+1}=\dfrac{1}{c^a+c^{e+a}+1}=\dfrac{1}{c^a+c^e c^a+1}\ \ \ (2).$ If we multiply $(2)$ by $\dfrac{c^{-a}}{c^{-a}}$ , then $(2)$ become $\dfrac{1}{c^a+c^{-s}+1}=\dfrac{1}{c^a+c^{e+a}+1}\dfrac{c^{-a}}{c^{-a}}=\dfrac{c^{-a}}{c^e+c^{-a}+1}\ \ \ (3).$ Summing up $(1)$ and $(3)$ , we have $f(c)=\dfrac{c^e+c^{-a}+1}{c^e+c^{-a}+1}=1.$ Hence $f'(c)=0$ at $c=\pi$ .

$\large f(c) = \dfrac{1}{c^s +c^{-e}+1} + \dfrac{1}{c^e+c^{-a}+1} + \dfrac{1}{c^a +c^{-s}+1}$ $\large f'(c) = \dfrac{-1 (s c^{s-1} - e c^{-e-1} )}{(c^s + c^{-e} +1)^2} + \dfrac{-1 (e c^{e-1} - a c^{-a-1} )}{(c^e+ c^{-a} +1)^2} + \dfrac{-1 (a c^{a-1} - s c^{-s-1} )}{(c^a + c^{-s} +1)^2}$ Because answer is independent of c,s,a
Assuming e = s = a = 0 $\large f'(\pi) = \dfrac{-1 (0 \times \pi^{0-1} - 0 \times \pi^{0-1})}{(\pi^0 + \pi^0 + 1)^2} + \dfrac{-1 (0 \times \pi^{0-1} - 0 \times \pi^{0-1})}{(\pi^0 + \pi^0 + 1)^2} +\dfrac{-1 (0 \times \pi^{0-1} - 0 \times \pi^{0-1})}{(\pi^0 + \pi^0 + 1)^2}$ $\large \color{#3D99F6}{ f'(\pi) = 0}$