1. Given differential equation

   $\frac{dy}{dx}+2y\tan x=\sin x, \qquad 0<x<\frac{\pi}{2}$

   with condition

   $y\left(\frac{\pi}{3}\right)=0.$

2. Solve the linear differential equation

   This is a linear first-order DE of the form

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

   where

   $P(x)=2\tan x, \qquad Q(x)=\sin x.$

   The integrating factor is

   $I.F.=e^{\int 2\tan x\,dx} = e^{2\int \tan x\,dx}.$

   Since

   $\int \tan x\,dx = -\ln(\cos x),$

   we get

   $I.F.=e^{-2\ln(\cos x)}=\sec^2 x.$

3. Multiply the DE by the integrating factor

   $\sec^2 x\frac{dy}{dx}+2y\tan x\sec^2 x=\sin x\sec^2 x.$

   The left side becomes

   $\frac{d}{dx}(y\sec^2 x)=\sin x\sec^2 x.$

4. Integrate both sides

   $y\sec^2 x=\int \sin x\sec^2 x\,dx + C.$

   Now,

   $\sin x\sec^2 x = \frac{\sin x}{\cos^2 x}=\tan x\sec x.$

   Therefore,

   $\int \tan x\sec x\,dx=\sec x.$

   So,

   $y\sec^2 x=\sec x + C.$

   Hence,

   $y=\frac{\sec x + C}{\sec^2 x}=\cos x + C\cos^2 x.$

5. Use the initial condition

   Given

   $y\left(\frac{\pi}{3}\right)=0.$

   Since

   $\cos\frac{\pi}{3}=\frac12,$

   we get

   $0=\frac12 + C\left(\frac12\right)^2=\frac12+\frac{C}{4}.$

   Therefore,

   $\frac{C}{4}=-\frac12 \implies C=-2.$

   Thus,

   $y=\cos x-2\cos^2 x.$

6. Find the maximum value

   Let

   $t=\cos x, \qquad 0<x<\frac{\pi}{2} \implies 0<t<1.$

   Then

   $y=t-2t^2.$

   This is a downward-opening parabola. Its maximum occurs at

   $\frac{dy}{dt}=1-4t=0 \implies t=\frac14.$

   Since $t=\frac14$ lies in $(0,1)$, it gives the maximum.

   Now,

   y_{\max}=\frac14-2\left(\frac14\right)^2=rac14-2\cdot\frac1{16}=rac14-\frac18=\frac18.

7. Check options

   • A: $\frac18$ ✅
   • B: $\frac34$ ❌
   • C: $\frac14$ ❌
   • D: $\frac38$ ❌

So the maximum value of $y(x)$ is

$\boxed{\frac18}.$