$x = {e^{{{\tan }^{ - 1}}\frac{{y - {x^2}}}{{{x^2}}}}}$
$\therefore$ $\log x = {\tan ^{ - 1}}\frac{{y - {x^2}}}{{{x^2}}}$
$\therefore$ $\tan \left( {\log x} \right) = \frac{{y - {x^2}}}{{{x^2}}} = \frac{y}{{{x^2}}} - 1$
$\therefore$ $1 + \tan \left( {\log x} \right) = \frac{y}{{{x^2}}}$
$\therefore$ $y = {x^2}\left[ {\tan \left( {\log x} \right) + 1} \right]$
$\therefore$ $\frac{{dy}}{{dx}} = {x^2}\frac{d}{{dx}}\left[ {\tan \left( {\log x} \right) + 1} \right] + \left[ {\tan \left( {\log x} \right) + 1} \right]\frac{d}{{dx}}{x^2}$
$ = {x^2}{\sec ^2}\left( {\log x} \right).\frac{1}{x} + \left[ {\tan \left( {\log x} \right) + 1} \right].2x$
$ = 2x\left( {\tan \left( {\log x} \right) + 1} \right) + x{\sec ^2}\left( {\log x} \right)$
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