Answer
$$\eqalign{
& {\text{Center}}\left( { - 2,3} \right) \cr
& {\text{Foci }}\left( { - 2,3 - \sqrt 5 } \right){\text{ and }}\left( { - 2,3 + \sqrt 5 } \right) \cr
& {\text{Vertex }}\left( { - 2,0} \right){\text{ and }}\left( { - 2,6} \right) \cr
& {\text{Eccentricity }}\frac{{\sqrt 5 }}{3} \cr} $$
Work Step by Step
$$\eqalign{
& 9{x^2} + 4{y^2} + 36x - 24y + 36 = 0 \cr
& {\text{Group terms}} \cr
& \left( {9{x^2} + 36x} \right) + \left( {4{y^2} - 24y} \right) = - 36 \cr
& 9\left( {{x^2} + 4x} \right) + 4\left( {{y^2} - 6y} \right) = - 36 \cr
& {\text{Completing the square}} \cr
& 9\left( {{x^2} + 4x + 4} \right) + 4\left( {{y^2} - 6y + 9} \right) = - 36 + 9\left( 4 \right) + 4\left( 9 \right) \cr
& 9{\left( {x + 2} \right)^2} + 4{\left( {y - 3} \right)^2} = 36 \cr
& {\text{Divide both sides of the equation by 36}} \cr
& \frac{{{{\left( {x + 2} \right)}^2}}}{4} + \frac{{{{\left( {y - 3} \right)}^2}}}{9} = 1 \cr
& {\text{The equation has the standard form}} \cr
& \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = 1,{\text{ }}a > b \cr
& {\text{Comparing}} \cr
& \frac{{{{\left( {x + 2} \right)}^2}}}{4} + \frac{{{{\left( {y - 3} \right)}^2}}}{9} = 1 \cr
& \to h = - 2,{\text{ }}k = 3,{\text{ }}a = 3,{\text{ }}b = 2,{\text{ }}c = \sqrt 5 \cr
& {\text{With}} \cr
& \underbrace {{\text{Vertex }}\left( {h,k - a} \right){\text{ and }}\left( {h,k + a} \right)}_ \Downarrow \cr
& {\text{Vertex }}\left( { - 2,3 - 3} \right){\text{ and }}\left( { - 2,3 + 3} \right) \cr
& {\text{Vertex }}\left( { - 2,0} \right){\text{ and }}\left( { - 2,6} \right) \cr
& \underbrace {{\text{Foci }}\left( {h,k - c} \right){\text{ and }}\left( {h,k + c} \right)}_ \Downarrow \cr
& {\text{Foci }}\left( { - 2,3 - \sqrt 5 } \right){\text{ and }}\left( { - 2,3 + \sqrt 5 } \right) \cr
& {\text{Eccentricity }}e = \frac{c}{a} = \frac{{\sqrt 5 }}{3} \cr
& {\text{Center}}\left( {h,k} \right) \to {\text{Center}}\left( { - 2,3} \right) \cr
& {\text{Graph:}} \cr} $$
