The solution of the differential equation $(1+y^2)+(x-e^{{\tan }^{-1}y})\frac{dy}{dx}=0$, is:

Rewrite as $\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{{\tan }^{-1}y}}{1+y^2}$.This is a linear differential equation of the form $\frac{dx}{dy}+Px=Q$.Integrating Factor $IF=e^{\int \frac{1}{1+y^2}dy}=e^{{\tan }^{-1}y}$.Solution: $x(IF)=\int Q(IF)dy+C$.
$$x{e}^{{\tan }^{-1}y}=\int \frac{e^{{\tan }^{-1}y}}{1+y^2}{e}^{{\tan }^{-1}y}dy$$
$$x{e}^{{\tan }^{-1}y}=\int \frac{e^{2{\tan }^{-1}y}}{1+y^2}dy$$
Let $t={\tan }^{-1}y⟹dt=\frac{1}{1+y^2}dy$.
$$x{e}^{{\tan }^{-1}y}=\int e^{2t}dt=\frac{e^{2t}}{2}+k=\frac{1}{2}{e}^{2{\tan }^{-1}y}+k$$
Rearranging to match option D logic: $x{e}^{{\tan }^{-1}y}=\frac{1}{2}{e}^{2{\tan }^{-1}y}+k$.Wait, looking at original options: Option D is correct.