$\mathrm{y}={\tan }^{-1} \left(\frac{\log \left(\frac{\mathrm{e}}{{\mathrm{x}}^2}\right)}{\log {\mathrm{ex}}^2}\right)+{\tan }^{-1} \left(\frac{3+2\log \mathrm{x}}{1-6\log \mathrm{x}}\right)$ then $\frac{\mathrm{dy}}{\mathrm{dx}}=$

The given equation is $\mathrm{y}={\tan }^{-1} \left(\frac{\log \left(\frac{\mathrm{e}}{{\mathrm{x}}^2}\right)}{\log {\mathrm{ex}}^2}\right)+{\tan }^{-1} \left(\frac{3+2\log \mathrm{x}}{1-6\log \mathrm{x}}\right)$

This can be written as 

$\begin{array}{rcl}\mathrm{y}& =& {\tan }^{-1} \left(\frac{\log e-\log \left(x^2\right) }{\log {\mathrm{ex}}^2}\right)+{\tan }^{-1} \left(\frac{3+2\log \mathrm{x}}{1-\left(3\right)\left(2\right)\log \mathrm{x}}\right)\\ & =& {\tan }^{-1}\left(\frac{1-2\log x}{1+2\log x}\right)+{\tan }^{-1}\left(3\right)+{\tan }^{-1}\left(2\log x\right)\\ & =& {\tan }^{-1}1-{\tan }^{-1}\left(2\log x\right)+{\tan }^{-1}\left(3\right)+{\tan }^{-1}\left(2\log x\right)\\ & =& {\tan }^{-1}1+{\tan }^{-1}3\end{array}$

Since it is constant, its derivative is zero