How to find the projection of $x^2+y^2+z^2+2xyz=1$ on $S^2$? The primary problem is to find the singular points of $g$ when $g$ defines on $S^2$ that $g:S^2 \rightarrow R^3,(x,y,z)\rightarrow(yz-x,zx-y,xy-z)$.
$Note:$The singular point in the problem is the point when the rank of $dg_{p}：T_p S^2 \rightarrow T_{g(p)}R^3$ is less than 2.
I tried to find the when the 3 determinant of 2-order minor of the Jacobian of $(y\sqrt{1-x^2-y^2}-x,\sqrt{1-x^2-y^2}-y,xy-\sqrt{1-x^2-y^2})$ are all $0$.
The equation is hard to solve.So I turned to find the point in $R^3 \setminus (0,0,0)$ whose rank is  less than 3.
It is equivalent to find when the determinat of matrix
$$
 \left[
 \begin{matrix}
   -1 & z & y \\
   z & -1 & x \\
   y & x & -1
  \end{matrix}
  \right] 
$$
is $0$.
And I get the equation $x^2+y^2+z^2+2xyz=1$
The next,as I think, is to find the projection of $x^2+y^2+z^2+2xyz=1$ on$S^2$ through the map $(x,y,z)\rightarrow(\dfrac{x}{\sqrt{x^2+y^2+z^2}},\dfrac{y}{\sqrt{x^2+y^2+z^2}} ,\dfrac{z}{\sqrt{x^2+y^2+z^2}})$.
The intersection of $x^2+y^2+z^2+2xyz=1$ and  $x^2+y^2+z^2=1$ is obvious a solution.
$x^2+y^2=1,z=0;y^2+z^2=1,x=0;x^2+z^2=1,y=0$
But I have no idea how to find the other projections.What should I do next?
 A: Let $f \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be defined by the same equation as $g$ so $f(x,y,z) = (yz - x, zx - y, xy - z)$ and $g = f|_{S^2}$. Let $p = (x_0,y_0,z_0) \in S^2$ (so that $x_0^2 + y_0^2 + z_0^2 = 1$). The tangent plane $T_{p}S^2$ can be naturally identified with the plane
$$ T_p S^2 = \{ (x,y,z)^T \in \mathbb{R}^3 \, | \, x \cdot x_0 + y \cdot y_0 + z \cdot z_0 = 0 \} \subseteq \mathbb{R}^3. $$
By calculation, we have
$$ df|_p = \begin{pmatrix} -1 & z_0 & y_0 \\
z_0 & -1 & x_0 \\
y_0 & x_0 & -1 \end{pmatrix}$$
and so
$$ \det(df|_p) = -1(1 - x_0^2) - z_0(-z_0 - x_0 y_0) + y_0(z_0 x_0 + y_0) \\
= -1 + x_0^2 + z_0^2 + x_0 y_0 z_0 + y_0 z_0 x_0 + y_0^2 = 2 x_0 y_0 z_0. $$
Thus, unless $x_0 y_0 z_0 = 0$, we have $\operatorname{rank} (df|_p) = 3$ and so $\operatorname{rank} (dg|_p) = \operatorname{rank} \left( df|_p \right)|_{T_p S^2} = 2$. We conclude that the only suspicious critical points are the points on $S^2$ satisfying $x_0 y_0 z_0 = 0$.
Note that if we permute the coordinates of $\mathbb{R}^3$ then the coordinates of $f$ gets permuted in the same way and so it is enough to analyze the case $z_0 = 0$. For such points,
$$ T_pS = \operatorname{span} \{ (0,0,1)^T, (-y_0, x_0, 0)^T \} $$
and
$$ df|_p \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 & 0 & y_0 \\
0 & -1 & x_0 \\
y_0 & x_0 & -1  \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} y_0 \\ x_0 \\ -1 \end{pmatrix}, \\
df|_p \begin{pmatrix} -y_0 \\ x_0 \\ 0 \end{pmatrix} =  \begin{pmatrix} -1 & 0 & y_0 \\
0 & -1 & x_0 \\
y_0 & x_0 & -1  \end{pmatrix} \begin{pmatrix} -y_0 \\ x_0 \\ 0 \end{pmatrix} = \begin{pmatrix} y_0 \\ x_0 \\ x_0^2 - y_0^2 \end{pmatrix}. $$
The vectors above will be linearly dependent if and only if $x_0^2 - y_0^2 = -1$. Together with $x_0^2 + y_0^2 = 1$, we see that the only solutions are $(0,1,0)^T,(0,-1,0)^T$. The other cases can be deduced by symmetry.
