Answer
$$\frac{{{{\left( {x - 2} \right)}^2}}}{{1/3}} + \frac{{{{\left( {y + 3} \right)}^2}}}{{1/2}} = 1$$
Work Step by Step
$$\eqalign{
& 3{x^2} + 2{y^2} - 12x + 12y + 29 = 0 \cr
& {\text{Group terms}} \cr
& \left( {3{x^2} - 12x} \right) + \left( {2{y^2} + 12y} \right) = - 29 \cr
& 3\left( {{x^2} - 4x} \right) + 2\left( {{y^2} + 6y} \right) = - 29 \cr
& {\text{Complete the square and factor}} \cr
& 3\left( {{x^2} - 4x + 4} \right) + 2\left( {{y^2} + 6y + 9} \right) = - 29 + 3\left( 4 \right) + 2\left( 9 \right) \cr
& 3{\left( {x - 2} \right)^2} + 2{\left( {y + 3} \right)^2} = 1 \cr
& \frac{{{{\left( {x - 2} \right)}^2}}}{{1/3}} + \frac{{{{\left( {y + 3} \right)}^2}}}{{1/2}} = 1 \cr
& {\text{The equation has the form}} \cr
& \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = 1,{\text{ }}a > b \cr
& \underbrace {\frac{{{{\left( {x - 2} \right)}^2}}}{{1/3}} + \frac{{{{\left( {y + 3} \right)}^2}}}{{1/2}} = 1}_ \Downarrow \cr
& h = 2,k = - 3,a = \frac{1}{{\sqrt 2 }},b = \frac{1}{{\sqrt 3 }} \cr
& c = \sqrt {\frac{1}{2} - \frac{1}{3}} = \frac{{\sqrt 6 }}{6} \cr
& {\text{With}} \cr
& {\text{Vertex }}\left( {h,k - a} \right){\text{ and }}\left( {h,k + a} \right) \cr
& {\text{Vertex }}\left( {2, - 3 - \frac{1}{{\sqrt 2 }}} \right){\text{ and }}\left( {2, - 3 + \frac{1}{{\sqrt 2 }}} \right) \cr
& {\text{Center}}\left( {h,k} \right) \to {\text{Center}}\left( {2, - 3} \right) \cr
& {\text{Foci }}\left( {2, - 3 - \frac{{\sqrt 6 }}{6}} \right){\text{ and }}\left( {2, - 3 + \frac{{\sqrt 6 }}{6}} \right) \cr
& {\text{Eccentricity }}e = \frac{c}{a} = \frac{{\sqrt 6 /6}}{{1/\sqrt 2 }} = \frac{{\sqrt 3 }}{3} \cr
& {\text{Graph}} \cr} $$
