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

$\frac{dy}{dx}-y\tan x=2x\sec x,$

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

$\therefore y.\cos x=\int 2xdx=x^2+c$

Now, $y\left(0\right)=0\Rightarrow c=0,\therefore y=x^2\sec x$

$\Rightarrow y\text{'}=2x\sec x+x^2\sec x\tan x$

Now, $y\left(\frac{\pi }{4}\right)=\frac{{\pi }^2}{16}\times \sqrt{2}=\frac{{\pi }^2}{8\sqrt{2}}$

$y\left(\frac{\pi }{3}\right)=\frac{{\pi }^2}{9}\times 2=\frac{2{\pi }^2}{9}$

$y\text{'}\left(\frac{\pi }{4}\right)=\frac{2\pi }{4}\times \sqrt{2}+\frac{{\pi }^2}{8\sqrt{2}}\times 1=\frac{{\pi }^2}{8\sqrt{2}}+\frac{\pi }{\sqrt{2}}$

$y\text{'}\left(\frac{\pi }{3}\right)=\frac{2\pi }{3}\times 2+\frac{2{\pi }^2}{9}\times \sqrt{3}=\frac{2{\pi }^2}{3\sqrt{3}}+\frac{4\pi }{3}$