Originally Posted by

**TheEmptySet** First we take the partial to find to vectors tangent to the surface.

$\displaystyle \vec r(u,v) = 5u^2 \vec i +(2u-v^2)\vec j+5v^2 \vec k$

$\displaystyle r_u(u,v)=10u \vec i + 2 \vec j $

$\displaystyle r_v(u,v)=-2v \vec j +10v \vec k$

$\displaystyle r_u(1,3)=10(1) \vec i + 2 \vec j =10 \vec i +2 \vec j$

$\displaystyle r_v(1,3)=-2(3) \vec j +10(3) \vec k=-6 \vec j + 30 \vec k$

Now we take the cross product of these two vectors to get a Normal vector to the surface.

Can you finish from here?

Good luck.