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?
On
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.
On
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.
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.