I know from check information in the task that $$\frac 1 \rho\cdot {\frac{\partial}{\partial \phi}}(\cos^2\phi\cdot \tanh z) - \frac \partial {\partial z} \left( \sin2\phi \cdot \ln \sqrt[{\Large\,\rho\,}] \frac{\cosh\rho}{\cosh z} \right)=0$$
After calculation $$\frac{1}{\rho} \cdot \frac\partial {\partial \phi}(\cos^2\phi\cdot \tanh z)= \frac{1}{\rho} \cdot 2 \cos\phi \sin\phi \cdot \tanh z= \frac{1}{\rho} \cdot \sin2\phi \cdot \tanh z $$
I have problem with the second derivative
\begin{align} & \frac \partial {\partial z} \left( \sin2\phi \cdot \ln \sqrt[{\Large\,\rho\,}]{\frac{\cosh\rho}{\cosh z}} \right) \\[10pt] = {} & \sin2\phi \cdot \frac 1 {\sqrt[{\Large\,\rho\,}]{\frac{\cosh\rho}{\cosh z}}} \cdot \frac 1 \rho \cdot \left(\frac{\cosh\rho}{\cosh z} \right)^{(1/\rho)-1} \cdot \sinh z \\[10pt] = {} & \frac 1 \rho \cdot \sin2\phi\cdot \frac{ \left(\frac{\cosh\rho}{\cosh z} \right)^{(1/\rho)-1}} {\left(\frac{\cosh\rho}{\cosh z}\right)^{1/\rho}} \cdot \sinh z \end{align}
As we can compare the line 2nd and 4th one I'm not getting the same result I don't know what I'm doing wrong
$$ \frac \partial {\partial z} \left( \sin2\phi \cdot \ln \sqrt[{\Large\,\rho\,}]\frac{\cosh\rho}{\cosh z} \right)= \frac 1 \rho \cdot \sin2\phi\cdot \left(\frac{\cosh z}{\cosh\rho}\right)\cdot \sinh z$$
The issue is when you took the derivative of $$\sqrt[\rho]\frac{\cosh \rho}{\cosh z}$$ Yor mistake is that $\frac{\partial}{\partial z}\frac{\cosh\rho}{\cosh z}\ne \sinh z$ but instead $$\frac{\partial}{\partial z}\frac{\cosh\rho}{\cosh z}=-\frac{\sinh z \cosh\rho}{\cosh^2 z}$$