I've noticed in my free time when the functional mapping $f(z+c)=-1/(f(z)+1)$ is iterated twice, it yields the original function $f(z)$ (i.e. $f(z+3c)=f(z)$). So I thought to study it as a periodic function...but I don't know enough about it to evaluate its Fourier series. I went back to its original equation, and saw that if it had a zero (I'll call it $\alpha$), it would also have a singularity at $\alpha-c$; I realized that likewise if $f(z)$ doesn't vanish it can't have any singularities (but I'm not sure that's possible since its equation cannot be satisfied by an exponential function).
I would like to know the most general kind of function $f(z)$ can be. It looks to me like a trigonometric function could describe it...but nothing appears to disqualify it having a second period. I am assuming $f(z)$ is meromorphic and continuous in the interval $(0,c)$, and I'm allowing $c$ to be complex (though not forcing it).
There was a mistake in my calculations and it now doesn't seem dolvable anymore for me, but I don't want to completely delete the andere.
First of all, let $x_0$ be a solution of $x^2 + x + 1 = 0$. We have \begin{align*} x^2+x+1=0 \iff x(x+1)=-1 \iff x = - \frac{1}{x+1}. \end{align*} Thus, $f(z) = x_0$ is a solution for your problem. Now \begin{align*} f'(z) = f'(z-c+c) = -(\frac{1}{f(z-c)+1})' = \frac{f'(z - c)}{(f(z-c)+1)^2}. \end{align*}