Analyzing Intervals of Increase and Decrease: 2^y = -(x + 1/6)

To solve this problem, we'll examine how the function behaves based on the structure given:

1. Rearrange the equation if needed: $2^y = -\left( x + \frac{1}{6} \right)$ implies: $y = \log_2 \left(-\left(x + \frac{1}{6}\right)\right)$. This hints $y$ is undefined unless $x < -\frac{1}{6}$; otherwise, the logarithm argument is non-positive.

2. Recognize: Derived behavior as $x \to -\frac{1}{6}^-$, $y$ shoots toward large negative values (approaches as $-\infty$).

3. Here, solving directly for a derivative doesn't computationally proceed without explicit form, but relies on boundary behavior.

4. Therefore, examine if behavior up to $x = -\frac{1}{6}$ makes it decrease (progressively smaller $y$ as $x$ reduces), and afterwards (impossible), turns - thus indicating:

The function decreases relative to $x > -\frac{1}{6}$

There's no valid interval for $x < -\frac{1}{6}$.

Thus, the solution highlights these ranges:

Therefore, the intervals are given by:

$\searrow:x>-\frac{1}{6}\\\nearrow:x<-\frac{1}{6}$

Hence, the correct answer choice is:

4: $\searrow: x > -\frac{1}{6}$
$\nearrow: x < -\frac{1}{6}$