1. Given differential equation

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

   Rewrite it as

   $\left(x-e^{\tan^{-1}y}\right)\frac{dy}{dx}=-(1+y^2).$

2. Treat $x$ as a function of $y$

   Since $\dfrac{dy}{dx}$ appears, it is convenient to invert:

   $\frac{dx}{dy}=-\frac{x-e^{\tan^{-1}y}}{1+y^2}.$

   So,

   $\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{\tan^{-1}y}}{1+y^2}.$

   This is a linear differential equation in $x$ as a function of $y$.

3. Identify integrating factor

   Compare with

   $\frac{dx}{dy}+P(y)x=Q(y),$

   where

   $P(y)=\frac{1}{1+y^2}.$

   Therefore the integrating factor is

   $\text{I.F.}=e^{\int \frac{1}{1+y^2}\,dy}=e^{\tan^{-1}y}.$

4. Multiply throughout by the integrating factor

   =\frac{e^{2\tan^{-1}y}}{1+y^2}. $$ The left side becomes $$ \frac{d}{dy}\left(xe^{\tan^{-1}y}\right)=\frac{e^{2\tan^{-1}y}}{1+y^2}. $$

5. Integrate both sides

   $xe^{\tan^{-1}y}=\int \frac{e^{2\tan^{-1}y}}{1+y^2}\,dy + k.$

   Put

   $t=\tan^{-1}y \quad \Rightarrow \quad dt=\frac{dy}{1+y^2}.$

   Then the integral becomes

   $\int e^{2t}\,dt=\frac{1}{2}e^{2t}+k.$

   Hence,

   $xe^{\tan^{-1}y}=\frac{1}{2}e^{2\tan^{-1}y}+k.$

6. Simplify

   Multiply by $2$:

   $2xe^{\tan^{-1}y}=e^{2\tan^{-1}y}+k.$

7. Match with options

   This is exactly Option C.

Verification with stored answer: Stored correct answer is C, which matches the derived answer.