I have been trying to solve a differential equation as a practice question for my test, but I am just unable to get the correct answer. Please have a look at the D.E:  

$dy/dx = 1/(3x+\sin(3y))$ 

My working is as follows:  

$dx/dy = 3x+\sin(3y)$  

$dx - 3x = \sin(3y) dy$  

Integrating both sides:  

$x-(3/2)x^2 = -(1/3)\cos(3y) + c$

But the correct answer is:  

$x = ce^3y - 1/6(\cos(3y) + \sin(3y))$, which is quite different from what I have got. Could someone please help me solve this? Thanks :)