Let's assume you have some function $F = e^ab(c(t))$, where $a \in \mathbb{R}$ and $b(c(t))$ is a function, also $c(t)$ is a function.
To give the Euler equation of $c(t)$ we have to use: $$\frac{\partial F}{\partial c} - \frac{dF}{dt}(\frac{\partial F}{\partial \dot{c}}) = 0$$
The first element is equal to: $$ \frac{\partial F}{\partial c} = e^a\dot{b}(c(t))\dot{c}(t)$$
Now the question is what is the derivative of F w.r.t. $\dot{c}$, I think it is $0$ can anyone confirm this?
For a variational problem, the functional $F$ must be given in the form $F(\dot{c}, c, t)$. The actual dependence of $c$ on $t$ can't be part of its definition, because that's currently unknown. We only worry about the dependence of $c$ on $t$ once we start finding the differential equation for it, as it's needed to evaluate the total derivatives that appear in Euler-Lagrange.
So, to properly set up the problem, you'd want to write $$ F(\dot{c},c,t) = e^a b(c) $$ with the relevant Euler-Lagrange equation being $$ \frac{\partial F}{\partial c} - \frac{d}{dt}\left(\frac{\partial F}{\partial \dot{c}} \right)= 0 $$ (along with some boundary conditions set by whatever integral you're minimizing). Clearly $\partial_\dot{c} F = 0$, because the definition of $F$ above has no explicit dependence on $\dot{c}$. It does have explicit dependence on $c$, so we can use simple differentiation to get $\partial_c F = e^a b'(c)$.