Solution
= Solution
For the graph surface $Z=z(x,y)$, an upward normal is $(-z_x,-z_y,1)$. At $x=y=1$, let $q=e^{-1/2}$; then $z_x=q$ and $z_y=0$. Normalizing gives the <unit normal vector>
$$
\boxed{n=\frac{(-q,0,1)}{\sqrt{1+q^2}}=\frac{(-e^{-1/2},0,1)}{\sqrt{1+e^{-1}}}.}
$$
Its negative is the equally valid downward normal.