Answer
$${\left( {x - 3} \right)^2} = 8\left( {y + 1} \right)$$
Work Step by Step
$$\eqalign{
& {x^2} - 6x - 8y + 1 = 0 \cr
& \left( {{x^2} - 6x} \right) = 8y - 1 \cr
& {\text{Complete the square}} \cr
& \left( {{x^2} - 6x + 9} \right) = 8y - 1 + 9 \cr
& {\left( {x - 3} \right)^2} = 8y + 8 \cr
& {\left( {x - 3} \right)^2} = 8\left( {y + 1} \right) \cr
& {\text{The equation has the form }}{\left( {x - h} \right)^2} = 4p\left( {y - k} \right),{\text{ so}} \cr
& {\left( {x - 3} \right)^2} = 8\left( {y - \left( { - 1} \right)} \right) \Rightarrow h = 3,k = - 1 \cr
& {\text{The vertex is }}\left( {h,k} \right){\text{:}}\left( {3, - 1} \right) \cr
& {\text{Focus: }}\left( {h,p + k} \right):\left( {3,1} \right) \cr
& {\text{Directrix: }}y = - p + k,{\text{ }}y = - 3 \cr} $$
