Difficult time finding critical points using Lagrange The function is $f(x,y,z) = xyz$ on $x^2 + y^2 + z^2 = 1$. So I have:
$yz = 2x \lambda \\
xz = 2y \lambda \\
xy = 2z \lambda \\
x^2 + y^2 + z^2 = 1$
I guessed $x = \pm 1, y = 0, z = 0, \lambda = 0$ but apparently this isn't a critical point. There's many more that I can't seem to find using algebra either. I can't divide anything out because I can't divide by $0$, and I also can't factor anything. How can I find all the critical points?
 A: Your four equations are fine, but your conjectural conclusion is wrong, or at least incomplete. 
Multiplying the first three equations respectively by $x$, $y$, and $z$ and adding them up gives
$$3xyz=2\lambda(x^2+y^2+z^2)=2\lambda\ .$$
It follows that any point $(x,y,z)$ being a part of the solution would have to satisfy
$$yz=x\cdot 3xyz, \quad xz= y\cdot 3xyz,\quad xy= z\cdot 3xyz\ .$$
These three equations can be written as
$$(3x^2-1) yz=0, \quad (3y^2-1) xz=0,\quad (3z^2-1) xy=0\ .\tag{1}$$
When $yz\ne0$ then the first equation enforces $x=\pm{1\over\sqrt{3}}$, and from the other two equations it then follows that $y=\pm{1\over\sqrt{3}}, \>$  $z=\pm{1\over\sqrt{3}}$. Since all $8$ possible choices of signs lead to a point with $x^2+y^2+z^2=1$ we have to put all eight of them on our candidate list.
When $y=0$ then the first and the third equation $(1)$ are fulfilled, and from the second equation $(1)$ it then follows that $x=0$ or $z=0$. The condition $x^2+y^2+z^2=1$ then enforces $x=0$, $z=\pm 1$ or else $x=\pm1$, $z=0$. On account of symmetry it follows that we obtain $8$ more conditionally critical points, namely the points where the coordinate axes intersect the sphere $S^2$.
Our candidate list now comprises $16$ points, and all of them are conditionally critical points of $f$ on $S^2$. In order to find the $\max$ and the $\min$ of $f$ on $S^2$ you now have to compare the values of $f$ in these points.
A: Perhaps something like this: 
$$ yz = 2x \lambda |\times  x\\
xz = 2y \lambda |\times y \\
xy = 2z \lambda |\times z\\ $$
$$ xyz = 2x^2 \lambda\\
xyz = 2y^2 \lambda \\
xyz = 2z^2 \lambda \\ $$
Now we can something like this for example divide firs two equations and get: 
$$ x^2=y^2=z^2 $$
Now solve this step by step - consider octants for each case and you'll get a bunch of solutions I guess. Symmetrical equations always include something bit more than the ordinary ones. 
A: The definition of critical points are all points satisfying $\nabla f=0$, which is 
$\partial f /\partial x = 0$
$\partial f /\partial y = 0$
$\partial f /\partial z = 0$
Therefore, the critical points you want to find should satisfy the following nonlinear system:
$yz=0$
$xz=0$
$xy=0$
$x^2+y^2+z^2=1$
Clearly, there exist lots of critical points. There is no closed form solutions for this system. 
