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

We have,
                      $\frac{dy}{dx}-y\tan x=2x\sec x$
Given equation is linear differential equation.
So,  $IF=e^{\int Pdx}=e^{-\int \tan xdx}$
$=e^{\mathrm{logcos}x}=\cos x$
$\therefore$ Solution is 
$$\begin{gathered} y\cos x=\int 2x.\sec x.\cos xdx+C \\ y\cos x=\int 2xdx+C \\ y\cos x=x^2+C \\ y\left(0\right)=0\Rightarrow C=0 \\ \therefore y=x^2\sec x \end{gathered}$$
Hence,  $y\left(\frac{\pi }{4}\right)=\frac{{\pi }^2}{8\sqrt{2}},y\left(\frac{\pi }{3}\right)=\frac{2{\pi }^2}{9}$
$y\text{'}\left(x\right)=2x\sec x+x^2\sec x\tan x$
$$\begin{gathered} y\text{'}\left(\frac{\pi }{4}\right)=\frac{\pi }{\sqrt{2}}+\frac{{\pi }^2}{16}\sqrt{2} \\ =\frac{\pi }{\sqrt{2}}+\frac{{\pi }^2}{8\sqrt{2}}=\frac{9{\pi }^2}{8\sqrt{2}} \\ y\text{'}\left(\frac{\pi }{3}\right)=\frac{2\pi }{3}\times 2+\frac{{\pi }^2}{9}\times 2\sqrt{3}=\frac{4\pi }{3}+\frac{2\sqrt{3}{\pi }^2}{9} \\ y\text{'}\left(\frac{\pi }{3}\right)=\frac{4\pi }{3}+\frac{2{\pi }^2}{3\sqrt{3}} \end{gathered}$$