A curve passes through the point \(\left(1, \frac{\pi}{6}\right)\). Let the slope of the curve at each point \((x, y)\) be \(\frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0\), then, the equation of the curve is

\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y}{x}+\sec \left(\frac{y}{x}\right) \)
\( \text { Put } y=\mathrm{v} x \)
\( \therefore \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{v}+x \frac{\mathrm{dv}}{\mathrm{d} x} \)
\( \therefore \mathrm{v}+x \frac{\mathrm{dv}}{\mathrm{d} x}=\mathrm{v}+\sec \mathrm{v} \)
\(\Rightarrow \cos \mathrm{v} \mathrm{dv}=\frac{1}{x} \mathrm{~d} x\)
Integrating both sides, we get
\(\sin \mathrm{V}=\log (x)+\mathrm{c} \)
\( \Rightarrow \sin \left(\frac{y}{x}\right)=\log (x)+\mathrm{c}\)
The curve passes through \(\left(1, \frac{\pi}{6}\right)\).
\(
\sin \left(\frac{\pi}{6}\right)=\log (1)+c \Rightarrow c=\frac{1}{2}\)
\(\therefore \) Equation (ii) becomes
\(
\sin \left(\frac{y}{x}\right)=\log (x)+\frac{1}{2}\)