Answer
$$\frac{{{x^2}}}{9} - 3{y^2} = 1$$
Work Step by Step
$$\eqalign{
& {\text{vertices at }}\left( { - 3,0} \right),\left( {3,0} \right);{\text{ passing through the point }}\left( { - 6, - 1} \right) \cr
& {\text{The }}y{\text{ - coordinate in the vertices are the same, then the hyperbola}} \cr
& {\text{has the equation }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{With vertices at }}\left( { \pm a,0} \right) \cr
& {\text{vertices at }}\left( { \pm 3,0} \right) \to a = 3 \cr
& \cr
& \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& \frac{{{x^2}}}{{{{\left( 3 \right)}^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& \frac{{{x^2}}}{9} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{passing through the point }}\left( { - 6, - 1} \right) \cr
& \frac{{36}}{9} - \frac{1}{{{b^2}}} = 1 \cr
& {\text{Solving for }}{b^2} \cr
& \frac{1}{{{b^2}}} = 3 \cr
& {b^2} = \frac{1}{3} \cr
& {\text{The equation of the hyperbola is}} \cr
& \frac{{{x^2}}}{{{{\left( 3 \right)}^2}}} - \frac{{{y^2}}}{{1/3}} = 1 \cr
& \frac{{{x^2}}}{9} - 3{y^2} = 1 \cr} $$
