The equation of the surface is given implicitly as:
$\ x^2yz +3y^2-2xz^2+8^z=0$
Am I supposed to simply calculate the gradient vector for this surface and then insert the values of my point ( which is $\ (1,2,-1)$ by the way )? If yes, I would like to understand the theoretical background of it (why it is that way).
On
Yes. Take the gradient and put in your points to get the orthogonal vector.
Now, this works because the gradient actually returns the plane tangent to the surface, which, like all planes, is defined by the normal vector and a point.
More specifically, remember that the function for a plane is: $$(\left<x,y,z\right>-\left<x_0,y_0,z_0\right>)\cdot\vec n=0$$
Recall that the equation of a line only requires that we have a point and a parallel vector. Since we want a line that is at the point $(x_0,y_0,z_0) = (1,2,-1)$ we know that this point must also be on the line and we know that $\triangledown f(1,2,-1) $ is a vector that is normal to the surface and hence will be parallel to the line. Therefore the equation of the normal line is, $$r(t) = \langle 1,2,-1\rangle + t \triangledown f(1,2,-1)$$