A plane is horisontal when $z = a$, where $a$ is a constant. Which values for $a$ makes $z = a$ describe a plane tangent to the surface given by $$ z = xye^{-\frac{9x^2+8y^2}{2}+1} ? $$
I computed the partial derivatives in the x- and y direction, and got that they are $0$ for $y = 0, x = 0, x= \pm\frac {1}{3} $
Which only gives me the solution $z = 0$, which is not correct. (i.e $z = 0$ is not the only horisontal plane tangent to the surface).
Consider the function $$f(x,y):=\bigl(xe^{-9x^2/2}\bigr)(ye^{-8y^2/2}\bigr)\ ,$$ and compute $$f_x=y(1-9x^2)\exp(\ldots),\qquad f_y=x(1-8y^2)\exp(\ldots)\ .$$ It follows that there are five critical points, namely $$(0,0),\quad\left(\pm{1\over3}, \pm{1\over2\sqrt{2}}\right)\ ,$$ whereby the $\pm$s can be chosen independently.