1. Given differential equation

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

   We are told that $x=f(y)$, so it is better to rewrite the equation in terms of $\dfrac{dx}{dy}$.

2. Convert to a linear differential equation in $x$ as a function of $y$

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

   Taking reciprocal form: $x-2e^{\tan^{-1}y}=-(1+y^2)\frac{dx}{dy}.$

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

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

3. Solve the linear ODE

   This is of the form $\frac{dx}{dy}+P(y)x=Q(y),$ where $P(y)=\frac{1}{1+y^2}.$

   Hence the integrating factor is $IF=e^{\int \frac{1}{1+y^2}\,dy}=e^{\tan^{-1}y}.$

4. Multiply through by the integrating factor

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

5. Integrate both sides

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

   Then $\int \frac{2e^{2\tan^{-1}y}}{1+y^2}\,dy = \int 2e^{2t}\,dt=e^{2t}+C=e^{2\tan^{-1}y}+C.$

   Therefore, $xe^{\tan^{-1}y}=e^{2\tan^{-1}y}+C.$

   So, $x=e^{\tan^{-1}y}+Ce^{-\tan^{-1}y}.$

6. Use the initial condition $f(0)=1$

   Since $x=f(y)$ and $f(0)=1$, at $y=0$ we have $x=1$.

   Also, $\tan^{-1}(0)=0.$

   Thus, $1=e^0+Ce^0=1+C \implies C=0.$

   Hence, $x=f(y)=e^{\tan^{-1}y}.$

7. Evaluate at $y=\dfrac{1}{\sqrt{3}}$

   $f\left(\frac{1}{\sqrt{3}}\right)=e^{\tan^{-1}(1/\sqrt{3})}.$

   Since $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6},$ we get $f\left(\frac{1}{\sqrt{3}}\right)=e^{\pi/6}.$

8. Option check

   • A: $e^{\pi/4}$ ❌
   • B: $e^{\pi/12}$ ❌
   • C: $e^{\pi/6}$ ✅
   • D: $e^{\pi/3}$ ❌

Therefore, the correct answer is Option C.