 Solution

The cylinder is a level set of $h=x^2+y^2-8$, so its outward normal vector is parallel to $\nabla h=(2x,2y,0)$. At the point this is $(4,4,0)$, with norm $4\sqrt{2}$. Choosing the outward orientation givesIts negative is the equally valid inward unit normal vector; the next directional derivative then changes sign.