If y(x) satisfies the differential equation $y^{\mathrm{\prime }}-y\tan x=2x\sec x$ and $y\left(0\right)=0$, then

We have,

$\frac{dy}{dx}-(\tan x)y=2x\sec x$                  ... (i)

It is a linear differential equation with

I.F.$=e^{-\int \tan xdx}=e^{\log \cos x}=\cos x$

Multiplying (i) by I.F. = cos x and integrating, we get

$\Rightarrow y\cos x=x^2+C$                       ... (ii)

Putting x = 0 and y = 0, we get  

$0=0+C\Rightarrow C=0$

Putting C = 0 in (ii), we get

$y=x^2\sec x$                                       ... (iii)

$\Rightarrow y^{\mathrm{\prime }}=2x\sec x+x^2\sec x\tan x$      ... (iv)

Putting $x=\frac{\pi }{3}$ in (iii) and (iv), we get

$y\left(\frac{\pi }{3}\right)=\frac{2{\pi }^2}{9},y^{\mathrm{\prime }}\left(\frac{\pi }{3}\right)=\frac{4\pi }{3}+\frac{2{\pi }^2}{3\sqrt{3}}$