I know that this is basic but I don't see it right now.
Let be a parametrization of a surface, in my case, $\mathbb{x}(u,v)=(\cos(v)\sin(u),\sin(v)\cos(u),\cos(u))$ where $u\in(0,\pi)$ and $v\in(-\pi,\pi)$.
How to see in a draw of the surface who are $x_{v}$ and $x_{u}$? The same for the $u$-parametric and $v$-parametric curves in the surface. Moreover, how to know the orientation of the normal vector? I mean, how to know if the normal vector $\frac{\mathbb{x_{u}}\times\mathbb{x_{v}}}{||\mathbb{x_{u}}\times\mathbb{x_{v}}||}$ points inwards or outwards the surface?
Thank you for the answers!
If you graph the function with a lattice showing the lines where either $u$ or $v$ is fixed and the other variable varies, then the partial vectors $X_u$ and $X_v$ will point in the same direction as the lines in the lattice. So if $\gamma_v(t)=x(t,v)$ is a parameterized curve where $v$ is fixed, then $X_u$ will be the tangent vector to the curve, and its magnitude will scale with the velocity of $\gamma.$ The same holds for $X_v$ and tangent vectors of curves where $u$ is fixed.
To determine the orientation of the normal vector, use the right hand rule. If you can wrap your right hand around through $X_u$ and then $X_v,$ the normal vector will point in the direction of your thumb. Determining orientation of a normal vector is a very visual task, and you will have to visualize the tangent vectors in order to do it.
To get you started on solving your specific problem, $$ X_u=\langle\cos v\cos u, -\sin v \sin u, -\sin u\rangle$$ and $$X_v=\langle -\sin v \sin u, \cos v \cos u, 0 \rangle.$$