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