I would like to get some information about the following functional equation:
$$f(z)=\frac{1}{24} - \frac{1}{2z} + \frac{\pi^2}{z^2} \left ( \frac{1}{6} -4 \ f \left (\frac{-4\pi^2}{z}\right )\right )$$
This functional relationship must hold only for $\mathfrak{R}(z)>0$. I would like to know wether it is possible to know what types of functions hold it.
To be clear, I do not know where to start from. I have barely worked with functional equations in my life.
Any help or bibliography will be welcomed.
Edit: Answering the comments, we are supposed to know that $f(z)$ as $z \to 0^+$ is $ O\left ( \frac{1}{z^2} \right )$. Moreover, we know that $f(-2\pi)=f(2\pi)=\frac{1}{24} - \frac{1}{8\pi}$. Thank you for the help received.
Edit 2: $f(z)$ is even: $f(z)=f(-z)$. From here, we could get the follwing functional equation for $\mathfrak{R}(z)<0$:
$$f(z)=\frac{1}{24} + \frac{1}{2z} + \frac{\pi^2}{z^2} \left ( \frac{1}{6} -4 \ f \left (\frac{4\pi^2}{z}\right )\right )$$
So that we can remove the $-$ sign from the first equation to get for $\mathfrak{R}(z)>0$:
$$f(z)=\frac{1}{24} - \frac{1}{2z} + \frac{\pi^2}{z^2} \left ( \frac{1}{6} -4 \ f \left (\frac{4\pi^2}{z}\right )\right )$$