Answer
$$y = 2{\text{ and }}12x + 3y - 2z = 12$$
Work Step by Step
$$\eqalign{
& {x^2} + \frac{{{y^2}}}{4} - \frac{{{z^2}}}{9} = 1;{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {0,2,0} \right){\text{ and }}\left( {1,1,\frac{3}{2}} \right) \cr
& {\text{Let }}F\left( {x,y,z} \right) = {x^2} + \frac{{{y^2}}}{4} - \frac{{{z^2}}}{9} - 1 \cr
& {\text{Calculate the partial derivatives }} \cr
& {F_x}\left( {x,y,z} \right){\text{, }}{F_y}\left( {x,y,z} \right){\text{and }}{F_z}\left( {x,y,z} \right) \cr
& {F_x}\left( {x,y,z} \right) = 2x \cr
& {F_y}\left( {x,y,z} \right) = \frac{y}{2} \cr
& {F_z}\left( {x,y,z} \right) = - \frac{{2z}}{9} \cr
& {\text{Evaluate at the point }}\left( {0,2,0} \right){\text{ and }}\left( {1,1,\frac{3}{2}} \right) \cr
& {\text{at }}{F_x}\left( {x,y,z} \right),{\mkern 1mu} {F_y}\left( {x,y,z} \right){\text{ and }}{F_z}\left( {x,y,z} \right) \cr
& {F_x}\left( {0,2,0} \right) = 0 \cr
& {F_y}\left( {0,2,0} \right) = 1 \cr
& {F_z}\left( {0,2,0} \right) = 0 \cr
& and \cr
& {F_x}\left( {1,1,\frac{3}{2}} \right) = 2 \cr
& {F_y}\left( {1,1,\frac{3}{2}} \right) = \frac{1}{2} \cr
& {F_z}\left( {1,1,\frac{3}{2}} \right) = - \frac{1}{3} \cr
& {\text{An equation of the plane tangent to the surface is}} \cr
& {F_x}\left( {a,b,c} \right)\left( {x - a} \right) + {F_y}\left( {a,b,c} \right)\left( {y - b} \right) + {F_z}\left( {a,b,c} \right)\left( {z - c} \right) = 0 \cr
& \cr
& {\text{*For the point }}\left( {0,2,0} \right) \cr
& 0\left( {x - 0} \right) + \left( {y - 2} \right) + 0\left( {z - 0} \right) = 0 \cr
& y - 2 = 0 \cr
& y = 2 \cr
& \cr
& {\text{*For the point }}\left( {1,1,\frac{3}{2}} \right) \cr
& 2\left( {x - 1} \right) + \frac{1}{2}\left( {y - 1} \right) - \frac{1}{3}\left( {z - \frac{3}{2}} \right) = 0 \cr
& 2x - 2 + \frac{1}{2}y - \frac{1}{2} - \frac{1}{3}z + \frac{1}{2} = 0 \cr
& 2x - 2 + \frac{1}{2}y - \frac{1}{3}z = 0 \cr
& 12x + 3y - 2z = 12 \cr
& \cr
& y = 2{\text{ and }}12x + 3y - 2z = 12 \cr} $$
