Solution:
\( x\cos x\, dy + (xy\sin x + y\cos x - 1)\, dx = 0 , \; 0 < x < \dfrac{\pi}{2} \)
\(\displaystyle \frac{dy}{dx} + \frac{(x\sin x + \cos x)}{x\cos x}\, y = \frac{1}{x\cos x} \)
\( I.F. = e^{\displaystyle\int \tan x + \frac{1}{x}dx} = e^{-\ln \cos x + \ln x} = \dfrac{x}{\cos x} \)
\(\displaystyle y \cdot \frac{x}{\cos x} = \tan x + c \)
\(\displaystyle \dfrac{3\sqrt{3}}{\pi} \cdot \dfrac{\dfrac{\pi}{3}}{\dfrac{1}{2}} = \sqrt{3} + c\)
\( c = \sqrt{3} \)
\(\displaystyle y \frac{x}{\cos x} = \tan x + \sqrt{3} \; ; \; y = \dfrac{\sin x}{x} + \frac{\sqrt{3}\cos x}{x} \)
\(\displaystyle y' = \frac{x(\cos x - \sqrt{3}\sin x) - (\sin x + \sqrt{3}\cos x)}{x^{2}} \)
\( \displaystyle y'\left(\dfrac{\pi}{6}\right) = \dfrac{\pi}{6}\left(\dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2}\right) - \dfrac{\left(\dfrac{1}{2} + \dfrac{3}{2}\right)}{\left(\dfrac{4}{5}\right)^2} = -\dfrac{2.36}{\pi^2} = -\dfrac{72}{\pi^2} \)
\( \dfrac{y''=x^{2}\!\left[\cos x - \sqrt{3}\sin x + x(-\sin x - \sqrt{3}\cos x) - (\cos x - \sqrt{3}\sin x)\right] - \left[x(\cos x - \sqrt{3}\sin x) + (\sin x + \sqrt{3}\cos x)2x\right]}{x^{4}} \)
\( \displaystyle \dfrac{y''\left(\dfrac{\pi}{6}\right)=\left(\dfrac{\pi}{6}\right)^2 \left(\dfrac{\pi}{6}\left(-\dfrac{1}{2} - \dfrac{3}{2}\right)\right) - \left(-\left(\dfrac{1}{2} + \dfrac{3}{2}\right)\right)\dfrac{\pi}{3}}{\left(\dfrac{\pi}{6}\right)^4} ; \; \left(\frac{\pi}{6}\right) y'\!\left(\frac{\pi}{6}\right) = -\frac{\pi^{3}}{3\cdot 6^{2}} \times \frac{6^{3}}{\pi^{3}} + \frac{2\pi}{3} \cdot \frac{6^{3}}{\pi^{3}} \)

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