isolate the x's and y's with parentheses and add 1 to both sides of the equation to get:
(y^2 - 6y) - (4x^2 - 8x) = 1
factor the 4 out of (4x^2 - 8x) to get:
(y^2 - 6y) - 4 * (x^2 - 2x) = 1
complete the squares on (y^2 - 6y) and (x^2 - 2x) to get:
(y-3)^2 - 9 - 4 * ((x-1)^2 - 1) = 1
simplify to get:
(y-3)^2 - 9 - 4*(x-1)^2 - 4*(-1) = 1
simplify further to get:
(y-3)^2 - 9 - 4(x-1)^2 + 4 = 1
add 9 and subtract 4 from both sides of the equation to get:
(y-3)^2 - 4(x-1)^2 = 1 + 9 - 4
simplify to get:
(y-3)^2 - 4(x-1)^2 = 6
divide both sides of the equation by 6 to get:
(y-3)^2 / 6 - 4(x-1)^2 / 6 = 1
multiply 4(x-1)^2 / 6 by (1/4) / (1/4) to get (x-1)^2 / (6/4) which makes your equation look like:
(y-3)^2 / 6 - (x-1)^2 / (6/4) = 1
the equation is now in the standard form of:
(y-k)^2 / a^2 - (x-h)^2 / b^2 = 1
you have a^2 = 6 which makes a = sqrt(6)
you have b^2 = 6/4 which makes b = sqrt(6) / 2
the equation for the asymptote for a vertically oriented hyperbola is:
y = plus or minus a/b (x-h)^2 + k
a/b = sqrt(6) / (sqrt(6) / 2) which becomes a/b = 2
h is equal to 1 and k is equal to 3
the equation for the asymptote becomes:
y = plus or minus 2(x-1) + 3
the graph of your hyperbola is shown below:
here's an excellent reference to help you understand how to work with hyperbolas.
http://www.purplemath.com/modules/hyperbola.htm