equations

D.E.

$ydx = (e^{3x}+1)dy$

Since the equation has separable variables,

$$\begin{eqnarray*} ydx &=& (e^{3x}+1)dy\\ \dfrac{dx}{e^{3x}+1} - \dfrac{dy}{y} &=& 0\\ \int \dfrac{e^{3x} dx}{e^{6x}+e^{3x}} - \int \dfrac{dy}{y} &=& \int 0\\ \dfrac{1}{3} (3x - \ln(e^{3x}+1)) - \ln y &=& C\\ 3x - \ln(e^{3x}+1) - 3\ln y &=& C\\ \end{eqnarray*}$$

Rearranging, the answer is $\boxed{e^{3x}=Cy^3(e^{3x}+1)}$