This is a problem from one of past exams from my course.
Consider a function $z(x, y)=\cos{x}e^{-x^2-y^2}$.
Show that stationary points of $z$ lie along the $x$-axis, and satisfy $\tan{x}=k(x)$. Give an expression for $k(x)$.
This one I managed to do, I got that for $z_x=z_y=0$ we have $y=0$ and $\tan{x}=-2x$.
Determine the nature of these stationary points of $z(x,y)$.
This one I have problems with. I know the conditions for maxima, minima etc. in two variables ($D=z_{xx}z_{yy}-z_{xy}^2$ test), but I got into some horrible algebraic mess when trying to determine $D$.
Is there any "cleverer" method to determine the nature of the stationary points?
For what it's worth, $z_x = -e^{-x^2-y^2}(2x\cos{x}+\sin{x})$ and $z_y = -2y\cos{x}e^{-x^2-y^2}.$
At any stationary point $(x,0)$ we have $$\eqalign{ \left[\matrix{z_{xx}&z_{xy}\cr z_{xy}&z_{yy}}{}\right]&= \left[ \begin{array}{cc} e^{-x^2} \left(\left(4 x^2-3\right) \cos (x)+4 x \sin (x)\right) & 0 \\ 0 & -2 e^{-x^2} \cos (x) \\ \end{array} \right]\cr &=-\cos(x)e^{-x^2}\left[\matrix{3+4x^2&0\cr0&2}\right] } $$ where in the last step, I used $\tan(x)=-2x$. Now, we have a local minimum on every stationary point $(x,0)$ with $\tan(x)=-2x$ and $\cos(x)<0$ and a local maximum on every stationary point $(x,0)$ with $\tan(x)=-2x$ and $\cos(x)>0$.