Find equations of the two lines through the origin that make an angle $\tan^{-1}(1/2)$ with $3y=2x$ 
If two straight lines pass through the origin and makes an angle $\tan^{-1}(1/2)$ with $3y=2x$, then find its equations.

Let $m$ be the gradient of the line then, $$\frac{1}{2}=\frac{m-2/3}{1-2m/3}$$ I don't know whether my approach is correct or not. 

It may be better to compute the gradient $m$ directly.
The angle between the line to be solved $y=mx$ and the $x$-axis is $\tan^{-1}(1/2)+\tan^{-1}(2/3)$. So,
$$m=\tan \left[ \tan^{-1}(1/2)+\tan^{-1}(2/3) \right]=\frac{1/2+2/3}{1-1/3}=\frac{7}{4}$$
If you solve your equation $1/2=(m-2/3)/(1+2m/3)$, you will get the same $m$. 
Thus, the line is,
$$y= \frac{7}{4} x $$
The gradient of the second line can be computed similarly and the line is $y=x/8$.
