1. Write the differential equation

Given slope: $\frac{dy}{dx}=\frac{y}{x}+\sec\left(\frac{y}{x}\right), \qquad x>0$

This is a homogeneous-type differential equation because the right-hand side is a function of $\dfrac{y}{x}$.

2. Use the substitution

Let $v=\frac{y}{x} \implies y=vx$

Then $\frac{dy}{dx}=v+x\frac{dv}{dx}$

Substitute into the differential equation: $v+x\frac{dv}{dx}=v+\sec v$

So, $x\frac{dv}{dx}=\sec v$

Hence, $\frac{dv}{dx}=\frac{\sec v}{x}$

or $\cos v\, dv=\frac{dx}{x}$

3. Integrate both sides

$\int \cos v\, dv=\int \frac{dx}{x}$

This gives $\sin v=\log x + C$

Now substitute back $v=\dfrac{y}{x}$: $\sin\left(\frac{y}{x}\right)=\log x + C$

4. Use the given point

The curve passes through $\left(1,\frac{\pi}{6}\right)$

So, $x=1,\quad y=\frac{\pi}{6}$

Then $\frac{y}{x}=\frac{\pi}{6}$

Substitute into the solution: $\sin\left(\frac{\pi}{6}\right)=\log 1 + C$

$\frac{1}{2}=0+C$

So, $C=\frac{1}{2}$

Therefore the equation of the curve is $\boxed{\sin\left(\frac{y}{x}\right)=\log x+\frac{1}{2}}$

5. Match with the options

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

So the correct option is: $\boxed{A}$