I once studies hyperbolas ("conic sections") in a college analytic geometry course, but I can't remember the details. We learned the standard forms, asymptotes, foci, and stuff like that. Some hyperbolas open up & down, left & right, and some open along 45 degree angles. But since I can't remember, we'll just try to analyze this thing you have.
First, we'll see what happens at x = 3 and y = -1. For x = 3, we have (y+1)^2 = 16, so y = -1 +/- 4, or y equals 3 or -5. We have two points: (3, 3) and (3, -5)!
For y = -1, x is undefined (square root of a negative number).
Now let's multiply through by 16 and see what the equation looks like:
4(x-3)^2 - (y+1)^2 = -16
And we'll rewrite again to find out where the right side of the equation turns negative:
(y+1)^2 = 4(x-3)^2 + 16
The right side of this equation is always positive, so y is defined for all values of x.
Now let's turn it around the other way:
4(x-3)^2 = (y+1)^2 - 16
The right side of this equation goes negative whenever (y+1)^2 < 16. This inequality is satisfied whenever the absolute value of y + 1 is less than 4. That is,
y + 1 < 4 ==> y < 3 or -(y + 1) < 4 ==> y + 1 > -4 ==> y > -5
It looks like the expression is undefined for y in the open interval between negative 5 and positive 3. That's your answer. Just turn it around and say y must be less than or equal to negative 5, or greater than or equal to positive 3.
This hyperbola must open up and down, with vertices at (3, 3) and (3, -5), and an axis of symmetry at x = 3.
I hope I'm right. I did this one by the seat of my pants! Also, the Yahoo Answers spell checker said I spelled "vertices" wrong. I think the spell checker needs to go back to school!