Finding points where a smooth map between differential manifolds is or is not an immersion. I am having trouble answering questions pertaining to immersions on smooth manifolds.  For example:  Given the unit sphere $S^2$ around the origin in $R^3$ and the map $f: S^2 \rightarrow R^3$ given by $f(x,y,z) =(yz,zx,xy), show f fails to be an immersion at 6 points.  
I am working on problems from a qualifying exam in topology and I am looking for a generic way to approach these types of problems.  
My ideas so far:
I know that an immersion at a point x in $S^2$ means that the derivative map $df_x$ is injective from the tangent space at that point in $S^2$ to the tangent space of f(x) in $R^3$.  
1)  Use stereographic projection and then consider the composition with this map giving us a map from $R^2$ to $R^3$.  This approach seems to be really ugly sometimes and if I replace $S^2$ with a hyperboloid or some other surface I may not have a map like stereographic projection.
2)  Consider only hemispheres at a time and solve for one variable.  In this case I would start by considering the top hemisphere and solve for z, getting $z = \sqrt{-x^2 -y^2 + 1}$ .  Then the $df_x$ map would just be the Jacobian of f with the above relation plugged in for z.  I can thus consider the rank of this map at different points to see if I have an immersion.  This method again gets pretty ugly and I am not even sure if it is correct. 
3)  The method I prefer but seem to get the opposite answer with is to consider the sphere as a level curve at 1 of the function $F(x,y,z) = x^2 + y^2 + z^2$.  Then to solve this problem I consider the 3x3 matrix for $df_x$.  If it is rank 3 then f is an immersion as it will take the tangent space at any point on the sphere to a two dimensional subspace of $R^3$.  If it is rank 2 then we need to show that the tangent space at a point is sent to a two dimensional space under $df_x$.  One way I believe we can do this is to consider the normal vector (gradient of F) to this tangent plane and see what happens to it under $df_x$.
IF the normal vector is in the kernel of $df_x$ then is it true that the tangent space is preserved?  Trying this method I got the answer that the points $(\pm 1, 0,0), (0, \pm1,0), (0,0, \pm1)$ ARE points at which we have an immersion and every other point on the great circles (z=0, x=0,y=0) are NOT immersions.  I know this is possible as an answer but seems to be the opposite of what the question asks.  
If anybody can give me any ideas on which method here works or any other generic method to do these problems I would be very appreciative.  Thanks.
Also this is my first question posted here so please let me know of any positing errors or anything I might have messed up.
 A: There is a clash of variables in your problem, $x$ is both a coordinate and a point. I am going to use $p$ as the point, with coordinates $p=(a,b,c)$.
Your method 3) starts out OK, but then you go off on the wrong track in how you use the gradient vector of $F$ at $p$. Your goal is to determine at what points $p$ the restricted map $df_p \mid T_p S^2$ is not an injection. The tangent plane $T_p S^2$ is the perpendicular to the gradient vector of $F$ at $p$; equivalently, but perhaps more useful, is that the tangent plane is the kernel of $dF_p$. So, you simply need to find those points $p$ such that the kernel of $df_p$ intersected with kernel of $dF_p$ is nontrivial. This reduces the problem to finding those points $p$ at which a certain system of homogeneous linear equations (whose coefficients depend on the coordinates of $p=(a,b,c)$) has a nontrivial solution, which is a nice concrete linear algebra problem.
A: There are not $6$, but $12$ such points.
Consider $f$ as a map ${\mathbb R}^3\to{\mathbb R}^3$. One computes
$$df(x,y,z)=\left[\matrix{0&z&y\cr z&0&x\cr y&x&0\cr}\right]\ .$$
This shows that $df$ is singular on the three coordinate planes. By symmetry it is enough to consider the plane $z=0$, in particular the points $p=(\cos\phi,\sin\phi,0)\in S^2$. Such a point $p$ is a singular point of $f\restriction S^2$ iff the kernel $K(p)$ of $df(p)$  lies in the tangent plane $T_p(S^2)$, or what is the same thing: iff $K(p)$ is orthogonal to $p$. Now $K(p)$ is spanned by the vector $(\cos\phi,-\sin\phi,0)$, so that we obtain the condition
$$(\cos\phi,\sin\phi,0)\cdot(\cos\phi,-\sin\phi,0)=0\ ,$$
which amounts to $\cos(2\phi)=0$. This produces the four $\phi$-values $\pm{\pi\over4}$,$\ \pm{3\pi\over4}$.
