Test whether the relation y = -sqrt(-5 - 6x - x²) is a function or not? If it is a function, determine its domain and range.

A relation is a function if for every input $x$ in its domain, there is exactly one output $y$.
The given relation is $y = -\sqrt{-5 - 6x - x^2}$.
For any valid value of $x$, the expression under the square root, $-5 - 6x - x^2$, will evaluate to a single non-negative number. The square root of this number, $\sqrt{-5 - 6x - x^2}$, will yield a single non-negative value (the principal square root). The negative sign in front, $-\sqrt{-5 - 6x - x^2}$, ensures that for each valid $x$, there is only one unique $y$ value.
Therefore, the relation is a function.

For the expression under the square root to be defined in real numbers, it must be non-negative:

Multiply the inequality by $-1$ and reverse the inequality sign:

Factor the quadratic expression:

The roots of the quadratic equation $x^2 + 6x + 5 = 0$ are $x = -1$ and $x = -5$.
Since the parabola $y = x^2 + 6x + 5$ opens upwards, the expression $(x+1)(x+5)$ is less than or equal to zero between its roots.
Thus, the domain is:

The domain is $\boxed{[-5, -1]}$.

The function is $y = -\sqrt{-5 - 6x - x^2}$.
Let's complete the square for the expression under the square root:

So, the function can be written as:

From the domain, we know that $-5 \le x \le -1$.
Let's find the range of $(x+3)^2$:

When we square this interval, the minimum value of $(x+3)^2$ is 0 (when $x+3=0$, i.e., $x=-3$) and the maximum value is 4 (when $x+3=-2$ or $x+3=2$).
So, $0 \le (x+3)^2 \le 4$.

Now, consider the expression under the square root, $4 - (x+3)^2$:
Since $0 \le (x+3)^2 \le 4$, we have:

This means the value inside the square root ranges from 0 to 4.
Next, consider $\sqrt{4 - (x+3)^2}$:

Finally, consider $y = -\sqrt{4 - (x+3)^2}$:
Multiplying by $-1$ reverses the inequality signs:

So, the range of the function is:

The range is $\boxed{[-2, 0]}$.