"Find all horizontal planes tangent to the surface given by $\displaystyle z=xye^\frac{-(x^2+y^2)}{2} $ or $\displaystyle f=-z + xye^\frac{-(x^2+y^2)}{2} $."

I already did this problem but there's something that bugs me. Here's how I solved it:

The normal vector to a curve at a point is $\displaystyle <f_x, f_y, f_z>$

If this curve is to be horizontal, $\displaystyle f_x $ and $\displaystyle f_y $ must be zero, and the only component is in the z direction.

So, $\displaystyle \frac{\partial f}{\partial x} = (xy)' e^\frac{-(x^2+y^2)}{2} +xy(e^\frac{-(x^2+y^2)}{2})' = ye^\frac{-(x^2+y^2)}{2} - x^2ye^\frac{-(x^2+y^2)}{2} =0$ and this simplifies to $\displaystyle y(1-x^2)=0$

Similarly, by finding $\displaystyle \frac{\partial f}{\partial x} =0$ I get $\displaystyle x(1-y^2)=0 $

So now I get two equatons:

(1) $\displaystyle y(1-x^2)=0$

(2) $\displaystyle x(1-y^2)=0 $

By setting y=0 in (1), x=0 in (2).

I also get four other points $\displaystyle (\pm1, \pm1)$

I did a few calculations and found 3 (!) tangent planes: including one at (0,0) which doesn't make sense... I even plotted the surface on Maple and saw that there cannot be a tangent plane at (0,0) because it would intersect the curve at other points.

And THAT is what I don't get... How come I get (0,0) as a point that has a horizontal tangent plane from my equations? Or should I just disregard this point when solving the problem for the same reason that I stated?

And how would I know for sure if I didn't have Maple, for example?