Is the surface $z=\ln \cos x-\ln \cos y$ minimal?
I found the partial derivatives:
$$\frac{\partial z}{\partial x}=-\mathrm{tg}x,\frac{\partial z}{\partial y}=\mathrm{tg}y,\frac{\partial ^2z}{\partial x^2}=-\mathrm{sec}^2x,\frac{\partial ^2z}{\partial y^2}=\mathrm{sec}^2y$$
After that I used the formulas for the main curvature:
$$E=1+\left ( \frac{\partial z}{\partial x} \right )^2=1+\mathrm{tg}^2x=\mathrm{sec}^2x$$ $$F=\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}=-\mathrm{tg}x\mathrm{tg}y$$ $$G=1+\left ( \frac{\partial z}{\partial y} \right )^2=1+\mathrm{tg}^2y=\mathrm{tg}^2y$$ $$L=\frac{\partial ^2z}{\partial x^2}:E=-\frac{\mathrm{sec}^2x}{\mathrm{sec}^2x}=-1$$ $$M=\frac{\partial ^2z}{\partial x\partial y}:E=0, \; \frac{\partial ^2z}{\partial x\partial y}=0$$ $$N=\frac{\partial ^2z}{\partial y^2}:G=\frac{\mathrm{sec}^2y}{\mathrm{sec}^2y}=1$$
The average curvature is calculated as follows $$H=\frac{k_1+k_2}{2}\Rightarrow k_1,k_2=\frac{L+N}{2}\pm \sqrt{\left ( \frac{L-N}{2} \right )^2+M}$$
Did I use the formulas correctly? I feel that I made a mistake in the calculation or in the formula itself, maybe the formula for the average curvature is different?
You can use this formula for calculating the mean curvature:
$H = -\frac{1}{2}\frac{\left\|\frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\right\|}{\left\|\frac{\partial^2\mathbf{r}}{\partial r^2}\right\|} \nabla^2 r^2$
and you can let your functions be functions of the complex variable $z$.
Since your equation satisfies the Laplacian equation:
$\nabla^2 z = \frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 0$
you get that
$H = -\frac{1}{2}\frac{\left\|\frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\right\|}{\left\|\frac{\partial^2\mathbf{r}}{\partial r^2}\right\|}0 = 0$.
Since every surface that has zero mean curvature is a minimal surface, that must mean that your function's surface is a minimal surface.
Good luck!