= Solution
Orthogonal projection of a small surface patch onto the $xy$ plane multiplies its scalar area by $\cos\alpha=\mathbf n\cdot\mathbf k>0$. Hence $dx\,dy=\cos\alpha\,dS$ and $d\mathbf S=\mathbf n\,dx\,dy/\cos\alpha$. Since $\mathbf n=\mathbf u/|\mathbf u|$ and $\mathbf u\cdot\mathbf k=1$, one has $\cos\alpha=1/|\mathbf u|$. The <vector area element of an implicit surface> is thus
$$
\boxed{d\mathbf S=\mathbf u\,dx\,dy=\left(\frac{3x^2+y}{y},\frac{x+z}{y},1\right)dx\,dy.}
$$
The positive final component implements the specified upward orientation.
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