A curve passes through the point $(1, \pi / 6)$. 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

$\sin \left(\frac{y}{x}\right)=\log x+\frac{1}{2}$

It is given that

$\frac{d y}{d x}=\frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0$             ....(i)

Let $y=v x$. Then, $\frac{d y}{d x}=v+x \frac{d v}{x}$

Substituting these values in (i), we obtain

$v+x \frac{d v}{d x}=v+\sec v$

$\Rightarrow x \frac{d v}{d x}=\sec v$

$\Rightarrow \cos v d v=\frac{1}{x} d x$

Integrating both sides, we get

$\sin v=\log x+C$

$\Rightarrow \sin \left(\frac{y}{x}\right)=\log x+C$              ....(ii)

It passes through the point $\left(1, \frac{\pi}{6}\right)$.

∴ $\sin \frac{\pi}{6}=\log 1+C \Rightarrow C=\frac{1}{2}$

Putting $C=\frac{1}{2}$ in (ii), we obtain the equation of the curve as

$\sin \left(\frac{y}{x}\right)=\log x+\frac{1}{2}$