Answer
$${\text{Asymptotes}}:{\text{ }}y = \pm \frac{{1/3}}{1}\left( {x + 1} \right) - 3$$
Work Step by Step
$$\eqalign{
& {x^2} - 9{y^2} + 2x - 54y - 80 = 0 \cr
& {\text{Group terms}} \cr
& \left( {{x^2} + 2x} \right) - \left( {9{y^2} + 54y} \right) = 80 \cr
& \left( {{x^2} + 2x} \right) - 9\left( {{y^2} + 6y} \right) = 80 \cr
& {\text{Completing the square}} \cr
& \left( {{x^2} + 2x + 1} \right) - 9\left( {{y^2} + 6y + 9} \right) = 80 + 1 - 9\left( 9 \right) \cr
& {\left( {x + 1} \right)^2} - 9{\left( {y + 3} \right)^2} = 0 \cr
& {\left( {x + 1} \right)^2} - \frac{{{{\left( {y + 3} \right)}^2}}}{{1/9}} = 0 \cr
& {\text{Degenerate the hyperbola.}} \cr
& {\text{Therefore,}} \cr
& \underbrace {{{\left( {x + 1} \right)}^2} - \frac{{{{\left( {y + 3} \right)}^2}}}{{1/9}} = 0}_{\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 0} \cr
& h = - 1,{\text{ }}k = - 3,{\text{ }}a = 1,{\text{ }}b = 1/3 \cr
& \underbrace {{\text{Asymptotes}}:{\text{ }}y = \pm \frac{b}{a}\left( {x - h} \right) + k}_ \Downarrow \cr
& {\text{Asymptotes}}:{\text{ }}y = \pm \frac{{1/3}}{1}\left( {x + 1} \right) - 3 \cr
& y = \pm \frac{1}{3}\left( {x + 1} \right) - 3 \cr
& {\text{Degenerate the hyperbola, with two lines intersecting at }}\left( {h,k} \right) \cr
& \left( { - 1, - 3} \right) \cr
& {\text{Graph:}} \cr} $$
