Answer
$\frac{{{y^2}}}{9} - \frac{{{x^2}}}{{27}} = 1$
Work Step by Step
$$\eqalign{
& {\text{Center: }}\left( {0,0} \right){\text{, Vertex:}}\left( {\underbrace 0_x,3} \right){\text{, Focus:}}\left( {\underbrace 0_x,6} \right) \cr
& x{\text{ is constant with center at the origin}}{\text{, then the equation of }} \cr
& {\text{the hyperbola is }} \cr
& \frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1{\text{ }}\left( {\bf{1}} \right) \cr
& and \cr
& {\text{Vertices }}\left( {0, - a} \right){\text{ and }}\left( {0,a} \right) \cr
& {\text{Vertex:}}\left( {0,3} \right) \to a = 3 \cr
& {\text{Foci }}\left( {0, - c} \right){\text{ and }}\left( {0,c} \right) \cr
& {\text{Focus:}}\left( {0,6} \right) \to c = 6 \cr
& {\text{We have that }}{c^2} = {a^2} + {b^2},{\text{ then}} \cr
& {\left( 6 \right)^2} = {\left( 3 \right)^2} + {b^2} \cr
& {b^2} = 27 \cr
& {\text{Substituting }}{a^2} = 9{\text{ and }}{b^2} = 27{\text{ into }}\left( {\bf{1}} \right) \cr
& \frac{{{y^2}}}{9} - \frac{{{x^2}}}{{27}} = 1 \cr} $$
