Answer
$$e = \frac{1}{3}$$
Work Step by Step
$$\eqalign{
& {\text{From the graph we can see:}} \cr
& {\text{Focus }}\left( {3,0} \right){\text{ and the point }}\left( { - 3,8} \right) \cr
& {\text{The equation of the ellipse is }}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{Foci}}\left( { \pm c,0} \right) \to c = 3 \cr
& {c^2} = {a^2} - {b^2} \cr
& {a^2} - {b^2} = 9 \cr
& {b^2} = {a^2} - 9 \cr
& \cr
& {\text{Substitute the point }}\left( { - 3,8} \right){\text{ into }}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& \frac{9}{{{a^2}}} + \frac{{64}}{{{b^2}}} = 1 \cr
& {\text{Substitute }}{b^2} = {a^2} - 9 \cr
& \frac{9}{{{a^2}}} + \frac{{64}}{{{a^2} - 9}} = 1 \cr
& {\text{Solving for }}{a^2}{\text{ we obtain}} \cr
& {a^2} = 81 \cr
& a = 9 \cr
& {\text{Find the eccentricity}} \cr
& e = \frac{3}{9} \cr
& e = \frac{1}{3} \cr} $$
