Answer
$$\frac{{{{\left( {x - 3} \right)}^2}}}{5} + \frac{{{{\left( {y - 4} \right)}^2}}}{9} = 1$$
Work Step by Step
$$\eqalign{
& {\text{Vertices }}\left( {3,1} \right){\text{ and }}\left( {3,7} \right) \cr
& x{\text{ value is constant, then the orientation of the major axis}} \cr
& {\text{is vertical, the equation of the ellipse is in the form}} \cr
& \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = 1{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Vertices }}\left( {h,k \pm a} \right) \cr
& h = 3,{\text{ }}k + a = 7,{\text{ }}k - a = 1, \to k = 4,{\text{ }}a = 3 \cr
& {\text{Eccentricity}} = \frac{c}{a} \cr
& \frac{c}{a} = \frac{2}{3} \to c = \frac{2}{3}a = \frac{2}{3}\left( 3 \right) \cr
& c = 2 \cr
& {b^2} = {a^2} - {c^2} \cr
& {b^2} = 9 - 4 \cr
& {b^2} = 5 \cr
& {\text{Substituting the constants into }}\left( {\bf{1}} \right) \cr
& \frac{{{{\left( {x - 3} \right)}^2}}}{5} + \frac{{{{\left( {y - 4} \right)}^2}}}{9} = 1 \cr} $$
