Find orientation of manifold 
In $\mathbb{R}^3$ consider the the 2-dim manifold
  $$
M:=\left\{(x,y,z)\in\mathbb{R}^3:z=xy\right\}.
$$
  Let the orientation of $M$ be in such a manner that in the Point $(0,0,0)$ the vector $e_3=(0,0,1)$ is the positive orientated normal vector. How is the orientation of M given?

Hello, i do not know if I understand it right.
Do I have to find an orientated Atlas of $M$ that gives an orientation $\sigma$ to which the normal field is positive orientated?
Or something else?
 A: If $z=xy$ then the natural choice for parameters is just $(x,y)$ and $R(x,y) = (x,y,xy)$ parametrizes the surfaces. The normal vector field is given by $N(x,y) = R_x \times R_y$,
$$ R_x \times R_y = (1,0,y) \times (0,1,x) = \langle -y, -x, 1 \rangle $$
hence clearly $N(0,0)= \langle 0,0,1 \rangle$ which shows the standard parametrization does in fact have the desired orientation for $z=xy$. No need for atlas here, as you can easily see the normal vector field is well-behaved on the $xy$-plane so the current choice of coordinates will work well for whatever problem we face.

Notice that if we calculate the determinant in your viewpoint we obtain:
$$ det \left[ \begin{array}{ccc} -y & 1 & 0 \\ -x & 0 & 1 \\ 1 & y & x  \end{array} \right] =y^2+x^2+1 \neq 0$$
for all $(x,y) \in \mathbb{R}^2$. Of course, for $z=xy$ there should be a symmetry for exchange of $x,y$ in equations, this allowed my brother to immediately deduce that my original normal vector was calculated wrong. Now our view-points agree, the normal vector field $N(x,y)$ is defined on the whole surface $z=xy$ and the determinant maintains its positive sign throughout. Incidentally, we should normalize $N(x,y)$ if we want it to be a unit vector field, I didn't bother since that is not needed for our interest here.
