Given, $$\frac{d y}{d x}-y \tan x=2 x \sec x$$

Comparing with the liner differential form

$$\begin{gathered} \frac{d y}{d x}+\mathrm{P} y=\mathrm{Q} \\ \Rightarrow \mathrm{P}=-\tan x \text { and } \mathrm{Q}=2 x \sec x \end{gathered}$$

Now, Integrating factor (I.F) $$=e^{\int P d x}$$

$$\begin{aligned} & =e^{\int-\tan x d x} \\ & =e^{-\log \sec x} \\ & =\cos x \end{aligned}$$

$$\begin{aligned} & \text { The solution will be } y \text { (I.F) }=\int Q \text { (I.F) } d x \\ & \Rightarrow \quad y \cos x=\int 2 x \sec x \cdot \cos x d x \\ & \Rightarrow \quad y \cos x=2 \frac{x^2}{2}+c \\ & \Rightarrow \quad y=x^2 \sec x+c \sec x \\ \end{aligned}$$

Now, $$y(0)=0$$

$$\begin{aligned} & \Rightarrow \quad 0=0^2 \sec 0+\mathrm{c} \mathrm{sec} 0 \\ & \Rightarrow \quad c=0 \\ & \therefore \quad y=x^2 \sec x \\ & y^{\prime}=x^2 \sec x \cdot \tan x+2 x \sec x \quad \text{[using Product Rule in Differentiation]}\\ \end{aligned}$$

Now,

$$\begin{aligned} y\left(\frac{\pi}{4}\right) & =\left(\frac{\pi}{4}\right)^2 \sec \frac{\pi}{4}=\frac{\pi^2}{16} \sqrt{2}=\frac{\pi^2}{8 \sqrt{2}} \\ y^{\prime}\left(\frac{\pi}{4}\right) & =\left(\frac{\pi}{4}\right)^2 \sec \frac{\pi}{4} \tan \frac{\pi}{4}+2 \cdot \frac{\pi}{4} \sec \frac{\pi}{4} \\ & =\frac{\pi^2}{16} \cdot \sqrt{2} \cdot 1+\frac{\pi}{2} \cdot \sqrt{2} \\ & =\frac{\pi^2}{8 \sqrt{2}}+\frac{\pi}{\sqrt{2}} \end{aligned}$$

$$\begin{aligned} y\left(\frac{\pi}{3}\right) & =\left(\frac{\pi}{3}\right)^2 \sec \frac{\pi}{3} \\ & =\frac{\pi^2}{9} \cdot 2=\frac{2 \pi^2}{9} \\ y^{\prime}\left(\frac{\pi}{3}\right) & =\left(\frac{\pi}{3}\right)^2 \sec \frac{\pi}{3} \cdot \tan \frac{\pi}{3}+2 \cdot \frac{\pi}{3} \cdot \sec \frac{\pi}{3} \\ & =\frac{\pi^2}{9} \cdot 2 \cdot \sqrt{3}+\frac{2 \pi}{3} \cdot 2 \\ & =\frac{2 \pi^2}{3 \sqrt{3}}+\frac{4 \pi}{3} \end{aligned}$$