"A plane is horizontal when $z=a$, where a is some constant. For which values of a does $z=a$ describe a tangent plane for the surface
$z = x*y*e^{\frac{-2x^2+6y^2}{2}+1}$
I tried differentiating with respect to $x$ and $y$ and set these equal to $0$ to get
$\cfrac d{dx} = (2x^2-1)*y*(-e^{-x^2+3y^2+1}) = 0$
which has solutions $ x = \frac{1}{\sqrt{2}}$, $ x = \frac{-1}{\sqrt{2}}$, and $ y = 0$
$\cfrac d{dy} = x(6y^2+1)*e^{-x^2+3y^2+1}$ which has only $x=0$ as it's real solution.
I'm trying to find some $(x,y)$ I can plug into $z$ to get out my horizontal plane at $z=a$, but obviously I've overlooked something since $x=0$ and $y=0$ gives $z=0$.
What is my mistake?