Answer
$$\frac{{4{x^2}}}{{405}} + \frac{{4{y^2}}}{{729}} = 1$$
Work Step by Step
$$\eqalign{
& e = \frac{2}{3};\,\,\,{\text{foci at }}\left( {0, - 9} \right),\,\,\left( {0,9} \right) \cr
& {\text{The }}x{\text{ coordinate is the same, then the ellipse has the equation}} \cr
& \frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1\,\,\,\,\,\,\left( {{\text{where }}a > b} \right){\text{ and foci }}\left( {0, \pm c} \right) \cr
& {\text{,then}} \cr
& c = 9 \cr
& e = \frac{c}{a} = \frac{2}{3} \cr
& a = \frac{c}{3} = \frac{9}{{2/3}} = \frac{{27}}{2} \cr
& {b^2} = {a^2} - {c^2} \cr
& {b^2} = {\left( {\frac{{27}}{2}} \right)^2} - {9^2} \cr
& {b^2} = \frac{{405}}{4} \cr
& \cr
& \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{Substituting }}a{\text{ and }}b \cr
& \frac{{{x^2}}}{{\frac{{405}}{4}}} + \frac{{{y^2}}}{{\frac{{729}}{4}}} = 1 \cr
& \frac{{4{x^2}}}{{405}} + \frac{{4{y^2}}}{{729}} = 1 \cr} $$
