Given differential equation:

$$e^x \tan y \, dx + (1 - e^x) \sec^2 y \, dy = 0$$

We can rewrite it as:

$$e^x \tan y \, dx = - (1 - e^x) \sec^2 y \, dy$$

Dividing both sides by $(1 - e^x) \tan y$ (assuming $\tan y \neq 0$ and $1 - e^x \neq 0$), we get

$$\frac{e^x}{1 - e^x} \, dx = - \frac{\sec^2 y}{\tan y} \, dy$$

Now, integrate both sides:

$$\int \frac{e^x}{1 - e^x} \, dx = - \int \frac{\sec^2 y}{\tan y} \, dy$$

For the left integral, substitute $u = 1 - e^x$, so $du = -e^x \, dx$ or $e^x \, dx = -du$. Then

$$\int \frac{e^x}{1 - e^x} \, dx = \int \frac{-du}{u} = -\ln|u| + \text{constant} = -\ln|1 - e^x| + \text{constant}$$

For the right integral, note that $\frac{d}{dy}(\tan y) = \sec^2 y$, so

$$\int \frac{\sec^2 y}{\tan y} \, dy = \ln|\tan y| + \text{constant}$$

Thus, the right-hand side becomes $-\ln|\tan y| + \text{constant}$.

Equating the integrals and combining constants, we have

$$-\ln|1 - e^x| = -\ln|\tan y| + \ln C$$

where $\ln C$ is an arbitrary constant. Multiplying by $-1$ gives

$$\ln|1 - e^x| = \ln|\tan y| - \ln C = \ln\left|\frac{\tan y}{C}\right|$$

Exponentiating both sides yields

$$|1 - e^x| = \left|\frac{\tan y}{C}\right|$$

which can be written as $1 - e^x = K \tan y$, where $K$ is an arbitrary constant (absorbing signs). Equivalently,

$$\tan y = C (1 - e^x)$$

where $C$ is an arbitrary constant. Note that $C = 0$ gives $\tan y = 0$, i.e., $y = n\pi$, which also satisfy the original equation.

Hence, the general solution is:

$$\boxed{\tan y = C (1 - e^x)}$$