If y = y (x) is the solution of the differential equation $\left(1+{\mathrm{e}}^{2\mathrm{x}}\right)\frac{\mathrm{dy}}{\mathrm{dx}}+2\left(1+{\mathrm{y}}^2\right){\mathrm{e}}^{\mathrm{x}}=0\text{ and }\mathrm{y}\left(0\right)=0$then  $6\left({\mathrm{y}}^{\mathrm{\prime }}\left(0\right)+{\left(\mathrm{y}\left({\log }_{\mathrm{e}}\sqrt{3}\right)\right)}^2\right)$is equal to:

$\frac{\mathrm{dy}}{1+{\mathrm{y}}^2}+\frac{2{\mathrm{e}}^{\mathrm{x}}}{1+{\mathrm{e}}^{2\mathrm{x}}}\mathrm{dx}=0$(i)
on integration
$\begin{array}{l}{\tan }^{-1}\mathrm{y}+2{\tan }^{-1}{\mathrm{e}}^{\mathrm{x}}=\mathrm{c}\\ \because \mathrm{y}\left(0\right)=0\\ \text{ so, }\mathrm{C}=\frac{\mathrm{\pi }}{2}\Rightarrow {\tan }^{-1}\mathrm{y}+2{\tan }^{-1}{\mathrm{e}}^{\mathrm{x}}=\frac{\mathrm{\pi }}{4}\\ \text{ from eq.(i), }{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{x}=0}=-1\\ \arg \mathrm{y}(\ln \sqrt{3})=-\frac{1}{\sqrt{3}}\\ 6\left[{\mathrm{y}}^{\mathrm{\prime }}\left(0\right)+\left(\mathrm{y}(\ln \sqrt{3}{)}^2\right]=6\left[-1+\frac{1}{3}\right]=-4\right.\end{array}$