Question 695437
Something is wrong.
The midpoint between your vertices is (-3,3).
(Its coordinates are the averages of the coordinates of the vertices: {{{(0+(-6))/2=3)}}} and {{{(3+3)/2=3}}}).
That should be the center of the hyperbola, and the point where the asymptotes intersect.
However, you asymptotes intersect at (3,-3).
 
A hyperbola with axes parallel to the x-axis and y-axis would have an equation
with a difference of squares involving x, y, and the coordinates of the center (h,k) of the hyperbola.
The equation for a hyperbola with a horizontal transverse axis, opening left and right, like ) (, would look like
{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}} with asymptotes {{{y=(b/a)(x-h) +k}}} and {{{y= - (b/a)(x-h) +k}}} and vertices (h+a, k) and (h-a, k)
 
The equation for a hyperbola with a vertical transverse axis, would look like
{{{(y-k)^2/a^2-(x-h)^2/b^2=1}}} with asymptotes {{{y=(a/b)(x-h) +k}}} and {{{y= - (a/b)(x-h) +k}}} and vertices (h, a-k) and (h, a+k)

If your asymptotes are {{{y=-x}}} and {{{y=x-6}}}, with slopes {{{1}}} and {{{-1}}}, then {{{a=b}}}.
You should be able to find {{{a=b}}}, {{{h}}} and {{{k}}} from the coordinates of the vertices.
If your vertices are truly (0,3) and (-6,3), the center is (-3,3),
and {{{a=3}}}, the distance from the center to each of the vertices.