Find a unit normal vector at a point
Solution 1:
You have a surface $xy - z = 0$
$\nabla(xy-z)$ will be normal to the surface.
$\nabla(xy-z) = (y,x,-1)$
at the point $(2,3,6)$ gives $(3,2,-1)$
and normalize $(\frac {3}{\sqrt {14}},\frac {2}{\sqrt {14}}, -\frac {1}{\sqrt {14}})$
Looking at what you have:
$z = xy$
$\frac {\partial z}{\partial x} = y\\ \frac {\partial z}{\partial y} = x$
$(1,0,y)$ and $(0,1,x)$ lie in the plane parallel to the surface. Taking the cross product
$(0,1,x)\times(1,0,y) = (y,x,-1)$