A curve passes through the point 1 , π 6 . Let the slope of the curve at each point x , y be y x + sec…

Given slope at (x, y) is

$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\text{sec}\left(\frac{y}{x}\right)$......$\left(i\right)$
let $\frac{\text{y}}{\text{x}}=\text{t}\Rightarrow \text{y}=\text{xt}$ 

Differentiating this wrt $x$

$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{t}+\text{x}\frac{\text{dt}}{\text{dx}}$......$\left(ii\right)$

Now using equation $\left(i\right)$ and $\left(ii\right)$, we have
$\text{t}+\text{x}\frac{\text{dt}}{\text{dx}}=\text{t}+\text{sec}\left(\text{t}\right)$

$\Rightarrow \frac{dt}{\sec \left(t\right)}=\frac{dx}{x}$

Integrating both sides.
$\int \cos \left(t\right)\text{dt}=\int \frac{1}{\text{x}}\text{dx}$
$\sin \left(t\right)=\ln \left(x\right)+\text{c}$ 

Now Put value of $t$ in above equation.
$\text{sin}\left(\frac{y}{x}\right)=\ln \left(x\right)+\text{c}$.....$\left(iii\right)$
Given that curve passes through $\left(1\text{, }\frac{\mathrm{\pi }}{6}\right)$ so put in above equation.
$\text{sin}\left(\frac{\mathrm{\pi }}{6}\right)=\ln \left(1\right)+\text{c}\Rightarrow \text{c}=\frac{1}{2}$

Now, put the value of $c$ in equation $\left(iii\right)$
$\text{sin}\left(\frac{y}{x}\right)=\ln \left(x\right)+\frac{1}{2}$