If \(y=\tan ^{-1}\left[\frac{\log \left(\frac{\mathrm{e}}{\mathrm{x}^2}\right)}{\log \left(\mathrm{ex}^2\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log \mathrm{x}}{1-6 \log \mathrm{x}}\right]\), then \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\)

\(\begin{aligned} & y=\tan ^{-1}\left[\frac{\log \left(\frac{\mathrm{e}}{\mathrm{x}^2}\right)}{\log \left(\mathrm{ex}^2\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right] \\ & =\tan ^{-1}\left[\frac{\log \mathrm{e}-\log \mathrm{x}^2}{\log \mathrm{e}+\log \mathrm{x}^2}\right]+\tan ^{-1} 3+\tan ^{-1}(2 \log \mathrm{x}) \\ & =\tan ^{-1}\left[\frac{1-\log \mathrm{x}^2}{1+\log \mathrm{x}^2}\right]+\tan ^{-1} 3+\tan ^{-1}\left(\log \mathrm{x}^2\right) \\ & =\tan ^{-1}(1)-\tan ^{-1}\left(\log \mathrm{x}^2\right)+\tan ^{-1} 3+\tan ^{-1}\left(\log \mathrm{x}^2\right) \\ & =\tan ^{-1}(1)+\tan ^{-1}(3) \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=0 \Rightarrow \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=0\end{aligned}\)