Answer
$$\left( { - \frac{1}{4}, - \frac{3}{2}, - \frac{5}{4}} \right)$$
Work Step by Step
$$\eqalign{
& z = 4{x^2} + 4xy - 2{y^2} + 8x - 5y - 4 \cr
& 4{x^2} + 4xy - 2{y^2} + 8x - 5y - 4 - z = 0 \cr
& {\text{Consider}} \cr
& F\left( {x,y,z} \right) = 4{x^2} + 4xy - 2{y^2} + 8x - 5y - 4 - z \cr
& {\text{Calculating the partial derivatives}} \cr
& {F_x}\left( {x,y,z} \right) = 8x + 4y + 8 \cr
& {F_y}\left( {x,y,z} \right) = 4x - 4y - 5 \cr
& {F_z}\left( {x,y,z} \right) = - 1 \cr
& {\text{Find the gradient }}\nabla F\left( {x,y,z} \right) \cr
& \nabla F\left( {x,y,z} \right) = \left( {8x + 4y + 8} \right){\bf{i}} + \left( {4x - 4y - 5y} \right){\bf{j}} - {\bf{k}} \cr
& {\text{Find the point}}\left( {\text{s}} \right){\text{ on the surface at which the tangent plane }} \cr
& {\text{is horizontal}}. \cr
& 8x + 4y + 8 = 0 \cr
& 4x - 4y - 5 = 0 \cr
& {\text{Solving the equations simultaneously we obtain}} \cr
& x = - \frac{1}{4},{\text{ }}y = - \frac{3}{2} \cr
& z = 4{x^2} + 4xy - 2{y^2} + 8x - 5y - 4 \cr
& z = 4{\left( { - \frac{1}{4}} \right)^2} + 4\left( { - \frac{1}{4}} \right)\left( { - \frac{3}{2}} \right) - 2{\left( { - \frac{3}{2}} \right)^2} + 8\left( { - \frac{1}{4}} \right) - 5\left( { - \frac{3}{2}} \right) - 4 \cr
& z = - \frac{5}{4} \cr
& {\text{The point is}} \cr
& \left( { - \frac{1}{4}, - \frac{3}{2}, - \frac{5}{4}} \right) \cr} $$
