Find the center, vertices, foci, & asymptotes of the hyperbola

x^2-y^2-2x+6y-9=0

&

64y^2-36x^2-512y-360x-2180=0

& one more...

9y^2-16x^2-54y-128x-31=0

These are all similar in solution, so I will do the middle one, and you should be able to do the others in a similar manner.

64y^2 - 512y - 36x^2 - 360x = 2180
64(y^2 - 8y) - 36(x^2 + 10x) = 2180
64(y^2 - 8y + 16) - 36(x^2 + 10x + 25) = 2180 + 1024 - 900
64(y - 4)^2 - 36(x + 5)^2 = 2304

(y - 4)^2/36 - (x + 5)^2/64 = 1

center (-5, 4)
a^2 = 36 and b^2 = 64, so a = 6 and b = 8
vertices: (-5, 10) and (-5, -2)

(y - 4)^2/36 - (x + 5)^2/64 = 0
(y - 4)^2/36 = (x + 5)^2/64
(y - 4)^ = (36/64) (x + 5)^2
y - 4 = (± 3/4) (x + 5)
y = (± 3/4) (x + 5) + 4
y = (3/4) x + 15/4 + 4 = (3/4)x + 31/4 }
y = (-3/4) x - 15/4 + 4 = (-3/4)x + 1/4 }
Those are the equations of the asymptotes.