Given differential equation

$\frac{dy}{dx}+2y\tan x=\sin x, \qquad y\left(\frac{\pi}{3}\right)=0$

This is a linear first-order differential equation:

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

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

Find the integrating factor

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

Since $\int \tan x\,dx=-\ln|\cos x|,$ we get $\int 2\tan x\,dx=-2\ln|\cos x|=\ln(\sec^2 x).$

Hence, $\text{I.F.}=\sec^2 x.$

Multiply the equation 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:

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

Integrate both sides

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

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

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

Therefore,

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

so

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

Use the initial condition

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

Substitute $x=\frac{\pi}{3}$:

$0=\cos\frac{\pi}{3}+C\cos^2\frac{\pi}{3}$

$0=\frac12 + C\left(\frac14\right)$

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

Hence,

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

Find the maximum value of $y(x)$

Let $t=\cos x, \qquad -1\le t\le 1.$

Then $y=t-2t^2.$

This is a downward opening parabola in $t$.

Differentiate with respect to $t$:

$\frac{dy}{dt}=1-4t$

For maximum,

$1-4t=0 \implies t=\frac14$

Since $\frac14\in[-1,1]$, this gives the maximum.

Now,

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

Compare with options

The maximum value is

$\boxed{\frac18}$

So the correct option is A.