Yes, writing the surface as [itex]f(x,y,z)= z+xy= 2[/itex] gives the normal at any point [itex](x_0, y_0, z_0)[/itex] as [itex]\nabla f= y_0\vec{i}+ x_0\vec{j}+ \vec{k}[/itex]. In order to give a line that passes through the origin, that vector must be parallel to the "position" vector of the point: [itex]x_0\vec{i}+ y_0\vec{j}+ z_0\vec{k}[/itex] and so must satisfy [itex]y_0= x_0t[/itex], [itex]x_0= y_0t[/itex], and [itex]z_0= t[/itex] for some t. The two equations [itex]y_0= x_0t[/itex] and [itex]x_0= y_0t[/itex] give [itex]x_0^2= y_0^2[/itex] so that [itex]y_0=\pm x_0[/itex]. With the last, [itex]z_0= t[/itex] we must have [itex]t= z_0[/itex] and so [itex]y_0= x_0t= x_0z_0[/itex]. Putting those together, [itex]y_0= x_0= x_0z_0[/itex] would give [itex]z_0= 1[/itex] so that [itex]1= 2- x_0y_0[/itex] [itex]x_0y_0= 1[/itex] so that [itex]x_0= y_0= 1[/itex] or [itex]x_0= y_0= -1[/itex]. That is, two such points are (1, 1, 1) and (-1, -1, 1). Or [itex]y_0= -x_0= x_0z_0[/itex] so we must have [itex]z_0= -1[/itex] and then [itex]x_0y_0= 3[/itex] which is not possible since [itex]x_0[/itex] and [itex]y_0[/itex] are of different signs.
