Explanation

(a)
Given that, $x = \exp\left\{ \tan^{- 1}\left( \frac{y - x^{2}}{x^{2}} \right) \right\}$
Taking log on both sides, we get
$\log{x = tan^{- 1}\left( \frac{y - x^{2}}{x^{2}} \right)}$
$\Rightarrow \ \frac{y - x^{2}}{x^{2}} = \tan{(\log{x)}}$
$\Rightarrow y = x^{2}\tan{(\log{x) + x^{2}}}$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx} = 2x\tan{(\log{x) + x^{2}\frac{\sec^{2}{(\log{x)}}}{x} + 2x}}$
$\Rightarrow \frac{dy}{dx} = 2x\tan{(\log{x) + x\sec^{2}{(\log{x) + 2x}}}}$
$\Rightarrow \frac{dy}{dx} = 2x\lbrack 1 + \tan{(\log{x)\rbrack + x\sec^{2}{(\log{x)}}}}$