= Solution
The first <partial derivatives> are
$$
z_x=e^{-x^2-y^2}(-\sin x-2x\cos x),\qquad z_y=-2y\cos x\,e^{-x^2-y^2}.
$$
If $\cos x=0$, then $z_x=-\sin x\,e^{-x^2-y^2}\ne0$, so no <stationary point> is lost by excluding those values. The remaining condition $z_y=0$ forces $y=0$, and $z_x=0$ then gives
$$
\boxed{\tan x_s=-2x_s,\qquad k(x)=-2x}.
$$
There are infinitely many such points. On every tangent branch $(n\pi-\pi/2,n\pi+\pi/2)$, the <function> $\tan x+2x$ has <derivative> $\sec^2x+2>0$ and goes from $-\infty$ to $+\infty$. The <intermediate value theorem> and strict monotonicity therefore give exactly one root on every branch. The central root is $x_s=0$, and oddness pairs the other roots at opposite signs.
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