Answer
$$\left( {2,2, - 4} \right)$$
Work Step by Step
$$\eqalign{
& z = {x^2} - xy + {y^2} - 2x - 2y \cr
& {x^2} - xy + {y^2} - 2x - 2y - z = 0 \cr
& {\text{Consider}} \cr
& F\left( {x,y,z} \right) = {x^2} - xy + {y^2} - 2x - 2y - z \cr
& {\text{Calculating the partial derivatives}} \cr
& {F_x}\left( {x,y,z} \right) = 2x - y - 2 \cr
& {F_y}\left( {x,y,z} \right) = - x + 2y - 2 \cr
& {F_z}\left( {x,y,z} \right) = - 1 \cr
& {\text{Find the gradient }}\nabla F\left( {x,y,z} \right) \cr
& \nabla F\left( {x,y,z} \right) = \left( {2x - y - 2} \right){\bf{i}} + \left( { - x + 2y - 2} \right){\bf{j}} - {\bf{k}} \cr
& {\text{Find the point}}\left( {\text{s}} \right){\text{ on the surface at which the tangent plane }} \cr
& {\text{is horizontal}}. \cr
& 2x - y - 2 = 0 \cr
& - x + 2y - 2 = 0 \cr
& {\text{Solving the equations simultaneosly we obtain}} \cr
& x = 2,{\text{ }}y = 2 \cr
& z = 3{x^2} + 2{y^2} - 3x + 4y - 5 \cr
& z = {x^2} - xy + {y^2} - 2x - 2y \cr
& z = {\left( 2 \right)^2} - \left( 2 \right)\left( 2 \right) + {\left( 2 \right)^2} - 2\left( 2 \right) - 2\left( 2 \right) \cr
& z = - 4 \cr
& {\text{The point is}} \cr
& \left( {2,2, - 4} \right) \cr} $$