Find the equation of the directrix of the parabola $x = -\frac{1}{6} y^2.$
To find the equation of the directrix of a parabola, I need to recall the standard form of a parabola that opens to the left or right: $x = \frac{1}{4p} y^2$, where $p$ is the distance from the vertex to the focus and the directrix. Comparing this form with the given equation, I can see that $\frac{1}{4p} = -\frac{1}{6}$, so $p = -\frac{3}{2}.$ This means that the parabola opens to the left, and the focus is $p$ units to the left of the vertex. To find the vertex, I can complete the square for $y$: $x = -\frac{1}{6} (y^2 - 0y + 0)$, so the vertex is at $(0, 0)$. Therefore, the focus is at $(-\frac{3}{2}, 0)$, and the directrix is a vertical line $p$ units to the right of the vertex. The equation of the directrix is $x = \frac{3}{2}.$ 