Given, the slope of curve at $$(x,y)$$ is

$$\begin{aligned} & \frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0 . \\ \therefore \quad & \frac{d y}{d x}=\frac{y}{x}+\sec \left(\frac{y}{x}\right) \quad \text{... (i)} \end{aligned}$$

Put $$y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$$ in the equation (i)

$$\begin{aligned} & \Rightarrow \quad v+x \frac{d v}{d x}=\frac{v x}{x}+\sec \left(\frac{v x}{x}\right) \\ & \Rightarrow \quad v+x \frac{d v}{d x}=v+\sec (v) \\ & \Rightarrow \quad x \frac{d v}{d x}=\sec v \\ & \Rightarrow \quad \cos v d v=\frac{d x}{x} \\ & \Rightarrow \quad \int \cos v d v=\int \frac{d x}{x} \\ & \Rightarrow \quad \sin v=\ln x+c \\ & \Rightarrow \quad \sin \left(\frac{y}{x}\right)=\ln x+c \quad \text{... (i)} \end{aligned}$$

Put $$x=1$$ and $$y=\frac{\pi}{6}$$ in the equation (ii)

$$\begin{aligned} \Rightarrow & & \sin \frac{\pi}{6} & =\log 1+c \\ \Rightarrow & & \frac{1}{2} & =0+c \\ \Rightarrow & & c & =\frac{1}{2} \end{aligned}$$

Put $$c=\frac{1}{2}$$ in the equation (ii)

$$\Rightarrow \quad \sin \frac{y}{x}=\log x+\frac{1}{2}$$

Hints:

(i) the slope of a curve at $$(x, y)$$ is equal to $$\frac{d y}{d x}$$

(ii) For the solution of homogeneous differential equation $$\frac{d y}{d x}=f\left(\frac{y}{x}\right)$$ put $$y=v x$$ i.e. $$\frac{d y}{d x}=v+x \frac{d v}{d x}$$.