$\log \sqrt{x^2+y^2}=\tan ^{-1}\left(\frac{x}{y}\right)$
Differentiate w.r.t ' $x$ ' on both sides,
$$
\begin{gathered}
\frac{1}{\sqrt{x^2+y^2}} \frac{d}{d x}\left(\sqrt{x^2+y^2}\right)=\frac{1}{1+\left(\frac{x}{y}\right)^2} \cdot \frac{d}{d x}\left(\frac{x}{y}\right) \\
\frac{1}{\sqrt{x^2+y^2}} \cdot \frac{1}{2 \sqrt{x^2+y^2}} \cdot \frac{d}{d x}\left(x^2+y^2\right) \\
=\frac{1}{\frac{y^2+x^2}{y^2}} \frac{1 \cdot y-x \cdot y^{\prime}}{y^2} \\
\frac{1}{2\left(x^2+y^2\right)}\left(2 x+2 y y^{\prime}\right)=\frac{y^2}{\left(x^2+y^2\right)} \frac{y-x y^{\prime}}{y^2} \\
x+y \cdot y^{\prime}=y^2\left(\frac{y-x y^{\prime}}{y^2}\right)
\end{gathered}
$$

$$
\begin{aligned}
x+y y^{\prime} & =y-x y^{\prime} \\
y y^{\prime}+x y^{\prime} & =y-x \\
y^{\prime}(x+y) & =y-x \\
y^{\prime} & =\frac{y-x}{x+y} \\
\frac{d y}{d x} & =\frac{y-x}{y+x}
\end{aligned}
$$
Hence, option (1) is correct.