Find the intervals in which the function $f(x) = 6 - 9x - x^{2}$ is strictly increasing or strictly decreasing.

(N/A) Given function: $f(x) = 6 - 9x - x^{2}$.
Step $1$: Find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(6 - 9x - x^{2}) = -9 - 2x$.
Step $2$: Find the critical point by setting $f'(x) = 0$.
$-9 - 2x = 0 \implies 2x = -9 \implies x = -\frac{9}{2}$.
Step $3$: The point $x = -\frac{9}{2}$ divides the real line into two intervals: $(-\infty, -\frac{9}{2})$ and $(-\frac{9}{2}, \infty)$.
Step $4$: Test the intervals.
For $x \in (-\infty, -\frac{9}{2})$,choose $x = -5$. Then $f'(-5) = -9 - 2(-5) = -9 + 10 = 1 > 0$. Thus,$f(x)$ is strictly increasing on $(-\infty, -\frac{9}{2})$.
For $x \in (-\frac{9}{2}, \infty)$,choose $x = 0$. Then $f'(0) = -9 - 2(0) = -9 < 0$. Thus,$f(x)$ is strictly decreasing on $(-\frac{9}{2}, \infty)$.