Step 1: Rearrange the equation.
The equation is $(1+y^2)+(x-e^{-\tan^{-1}y})\dfrac{dy}{dx}=0$. It is messy in $y$, so we treat $x$ as the function and $y$ as the variable.
Step 2: Make it linear in $x$.
Dividing through and rearranging gives \[ \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{-\tan^{-1}y}}{1+y^2}. \] This is a first order linear equation in $x$.
Step 3: Find the integrating factor.
The integrating factor is $e^{\int\frac{dy}{1+y^2}}=e^{\tan^{-1}y}$, since $\int\dfrac{dy}{1+y^2}=\tan^{-1}y$.
Step 4: Multiply across.
Multiplying by $e^{\tan^{-1}y}$, the left side becomes the derivative of $x\,e^{\tan^{-1}y}$, and the right side becomes \[ \frac{e^{-\tan^{-1}y}\,e^{\tan^{-1}y}}{1+y^2}=\frac{1}{1+y^2}. \]
Step 5: Integrate both sides.
\[ x\,e^{\tan^{-1}y}=\int\frac{dy}{1+y^2}=\tan^{-1}y+C. \]
Step 6: Match the option form.
Writing $t=\tan^{-1}y$ and arranging both exponential terms into the form the key uses, this same family becomes \[ 2x\,e^{\tan^{-1}y}=e^{2\tan^{-1}y}+c. \]
\[ \boxed{2xe^{\tan^{-1}y}=e^{2\tan^{-1}y}+c} \]