Given, differential equation

$$ \begin{aligned} & (x \cos x) d y+(x y \sin x+y \cos x-1) d x=0,0 < x < \frac{\pi}{2} \\\\ & \Rightarrow \frac{d y}{d x}+\frac{x y \sin x+y \cos x-1}{x \cos x}=0 \\\\ & \Rightarrow \frac{d y}{d x}+\left(\frac{x y \sin x+y \cos x}{x \cos x}\right)=\frac{1}{x \cos x} \\\\ & \Rightarrow \frac{d y}{d x}+\left(\tan x+\frac{1}{x}\right) y=\frac{1}{x \cos x} \end{aligned} $$

Which is linear differential equation in the form of

$$ \begin{aligned} & \frac{d y}{d x}+P y=Q \\\\ & \therefore \mathrm{IF}=e^{\int\left(\tan x+\frac{1}{x}\right) d x}=e^{(\log \sec x+\log x)}=e^{\log (x \sec x)}=x \sec x \end{aligned} $$

$\therefore$ The general solution of the given differential equation

$$ \begin{array}{rlrl} y \cdot \mathrm{IF} =\int(Q \times \mathrm{IF}) d x+c \\\\ \Rightarrow y(x \sec x) =\int\left(\frac{1}{x \cos x} x \sec x\right) d x+c \\\\ \Rightarrow x y \sec x =\int \sec ^2 x d x+c \\\\ \Rightarrow x y \sec x =\tan x+c ........(i) \end{array} $$

Since, $\frac{\pi}{3} y\left(\frac{\pi}{3}\right)=\sqrt{3} \Rightarrow y\left(\frac{\pi}{3}\right)=\frac{3 \sqrt{3}}{\pi}$

$$ \begin{aligned} & \therefore \frac{\pi}{3}\left(\frac{3 \sqrt{3}}{\pi}\right) \sec \left(\frac{\pi}{3}\right)=\tan \left(\frac{\pi}{3}\right)+c \\\\ & \Rightarrow \sqrt{3}(2)=\sqrt{3}+c \\\\ & \begin{aligned} \Rightarrow & c=\sqrt{3} \end{aligned} \end{aligned} $$

On putting the value of $c$ in Eq. (i), we get

$$ \begin{aligned} &x y \sec x =\tan x+\sqrt{3} \\\\ &\Rightarrow y =\frac{1}{x}(\sin x+\sqrt{3} \cos x) \\\\ & y^{\prime} =\frac{1}{x}(\cos x-\sqrt{3} \sin x)-\frac{1}{x^2}(\sin x+\sqrt{3} \cos x) \end{aligned} $$

$\begin{aligned} & \text { and } y^{\prime \prime}=\frac{1}{x}(-\sin x-\sqrt{3} \cos x)-\frac{1}{x^2}(\cos x-\sqrt{3} \sin x) -\frac{1}{x^2}(\cos x-\sqrt{3} \sin x) \\ & -\frac{2}{x^3}(\cos x-\sqrt{3} \sin x) \\\\ & \therefore\left|\frac{\pi}{6} y^{\prime \prime}\left(\frac{\pi}{6}\right)+2 y^{\prime}\left(\frac{\pi}{6}\right)\right|=|-2|=2 \\\\ & \end{aligned}$