Given the differential equation

$$\left(e^y + 1\right) \cos x \, dx + e^y \sin x \, dy = 0,$$

we aim to find the value of $ e^{y\left(\frac{\pi}{6}\right)} $ given that the solution $ y(x) $ passes through the point $\left(\frac{\pi}{2}, 0\right)$.

First, we recognize that the differential equation can be rearranged as:

$$d\left(e^y \sin x\right) + \cos x \, dx = 0.$$

Integrating this expression, we obtain:

$$ e^y \sin x + \sin x = C,$$

where $ C $ is a constant. Given that the solution passes through the point $\left(\frac{\pi}{2}, 0\right)$, we substitute these values into the equation to find $ C $:

$$ e^0 \sin\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = C \Rightarrow 1 + 1 = C \Rightarrow C = 2.$$

Thus, the equation simplifies to:

$$e^y \sin x + \sin x = 2.$$

We now need to determine the value of $ e^{y\left(\frac{\pi}{6}\right)} $. Substituting $ x = \frac{\pi}{6} $ into the equation, we get:

$$ \left(e^{y\left(\frac{\pi}{6}\right)} + 1\right) \cdot \sin\left(\frac{\pi}{6}\right) = 2.$$

Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, the equation becomes:

$$ \left(e^{y\left(\frac{\pi}{6}\right)} + 1\right) \cdot \frac{1}{2} = 2 \Rightarrow e^{y\left(\frac{\pi}{6}\right)} + 1 = 4 \Rightarrow e^{y\left(\frac{\pi}{6}\right)} = 3.$$

Therefore, the value of $ e^{y\left(\frac{\pi}{6}\right)} $ is $ 3 $.