= Solution
For increments $h=x-x_s$ and $k=y-y_s$, the second-order multivariable <Taylor expansion> is
$$
z=z_s+(z_x)_s h+(z_y)_s k+\frac12\left[(z_{xx})_s h^2+2(z_{xy})_s hk+(z_{yy})_s k^2\right]+o(h^2+k^2).
$$
At a <stationary point> $(x_s,0)$, $\sin x_s=-2x_s\cos x_s$ and $z_s=\cos x_s e^{-x_s^2}$. Direct differentiation gives
$$
(z_{xx})_s=-(4x_s^2+3)z_s,\qquad(z_{xy})_s=0,\qquad(z_{yy})_s=-2z_s.
$$
Thus, since this particular <function> is analytic, the expansion can be written
$$
\boxed{z(x,y)=z_s\left[1-\frac{4x_s^2+3}{2}(x-x_s)^2-y^2\right]+O\!\left(((x-x_s)^2+y^2)^{3/2}\right)}.
$$
There are no linear terms because the <gradient> vanishes, and no mixed quadratic term because the mixed <derivative> vanishes on the axis.
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