$\displaystyle (x+2e^{3y})\frac{dy}{dx}=1$

$\displaystyle (x+2e^{3y})dy=dx$

$\displaystyle (x+2e$^{3y})dy-dx=0$

$\displaystyle X_{y}\neq Y_{x}$

$\displaystyle \frac{-X_{y}+Y_{x}}{X}=\frac{1}{-1}=-1$

$\displaystyle m'-m=0$

$\displaystyle m=ce^{-x}$

$\displaystyle (x+2e^{3y})e^{-x}dy-e^{-x}dx=0$

it gets much more complicated from here

is there any easier approach

?