jmc

Completing the square in $x$ and $y,$ we get $$9(x+3{)}^2-(y-5{)}^2=1,$$which we can write as $$\frac{(x+3{)}^2}{1/9}-\frac{(y-5{)}^2}{1}=1.$$Thus, $a^2=\frac{1}{9}$ and $b^2=1,$ so $a=\frac{1}{3}.$ Therefore, the distance between the vertices is $2a=\boxed{\frac{2}{3}}.$