The solution of the differential equation

$\left(1+y^2\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0$, is 

$2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+K$

We have,

$\left(1+y^2\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0$

$\Rightarrow \frac{d x}{d y}+\frac{x}{1+y^2}=\frac{e^{\tan ^{-1} y}}{1+y^2}$

This is a linear differential equation with

Integrating factor = $e^{\int \frac{1}{1+y^2} d y}=e^{\tan ^{-1} y}$

Multiplying (i) by integrating factor and integrating, we get

$x e^{\tan ^{-1} y}=\int \frac{e^{2 \tan ^{-1} y}}{1+y^2} d y$

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

$\Rightarrow x e^{\tan ^{-1} y}=\frac{1}{2} e^{2 \tan ^{-1} y}+C$

$\Rightarrow 2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+K$, where $K=2 C$