Solution by separation of variables dx+e^{3x}dy= 0.

Given differential equation is:

(1) $\begin{equation*} dx + e^{3x}dy = 0 \end{equation*}$

We can separate the variables as follows:

(2) $\begin{equation*} dy = -\frac{dx}{e^{3x}} \end{equation*}$

Now, we can integrate both sides of the equation:

(3) $\begin{equation*} \int dy = -\int \frac{dx}{e^{3x}} \end{equation*}$

Integrating the left-hand side, we get:

(4) $\begin{equation*} y = C_1 -\int \frac{dx}{e^{3x}} \end{equation*}$

where $C_1$ is the constant of integration.

For the right-hand side, we can use the substitution $u = 3x$ and $du/dx = 3$ to obtain:

(5) $\begin{equation*} \int \frac{dx}{e^{3x}} = \frac{1}{3}\int e^{-u}du = -\frac{1}{3}e^{-u} + C_2 \end{equation*}$

where $C_2$ is another constant of integration. Substituting back $u=3x$ , we have:

(6) $\begin{equation*} \int \frac{dx}{e^{3x}} = -\frac{1}{3}e^{-3x} + C_2 \end{equation*}$

Substituting this back into our original equation, we have:

(7) $\begin{equation*} y = C_1 +\frac{1}{3}e^{-3x} + C_2 \end{equation*}$

where $C_1$ and $C_2$ are constants of integration.

Therefore, the solution to the differential equation $dx + e^{3x}dy = 0$ is:

(8) $\begin{equation*} y = -\frac{1}{3}e^{-3x} + C \end{equation*}$

where $C = C_1 + C_2$ is the constant of integration.