$x=e^{\tan ^{-1}\left(\frac{y-x^2}{x^2}\right)}$, then $\frac{d y}{d x}$ is

$x\left(2+\frac{y^2}{x^4}-\frac{2 y}{x^2}\right)+\frac{2 y}{x}$

$x=e^{\tan ^{-1}\left(\left(y-x^2\right) / x^2\right)}$

$\log x=\tan ^{-1}\left(\frac{y}{x^2}-1\right) \Rightarrow \frac{y}{x^2}-1=\tan \log x$

$\frac{y}{x^2}=1+\tan \log x$

$y=x^2+x^2 \tan \log x$

$\frac{d y}{d x}=2 x+2 x \tan \log x+x \sec ^2(\log x) \times \frac{1}{x}$

$=2 x+2 x \tan \log x+x \sec ^2(\log x)$

$=\frac{2}{x}\left(x^2+x^2 \tan \log x\right)+x \sec ^2(\log x)=\frac{2 y}{x}+1+\tan ^2(\log x)$

$\Rightarrow \frac{2 y}{x}+x\left(1+\left(\frac{y}{x^2}-1\right)^2\right)=\frac{2 y}{x}+1+\frac{y^2}{x^4}+1-\frac{2 y}{x^2}$

$=\frac{x y^2}{x^4}-\frac{2 y x}{x^2}+\frac{2 y}{x}+2x$

$=\frac{y^2}{x^3}-\frac{2 y}{x}+\frac{2 y}{x}+2x=\frac{y^2}{x^3}+2x$

Hence (1) is correct answer.