<s>[INST] Solving system of non-linear equations. So I'm trying to find the stationary points for $$f(x,y,z) = 4x^2 + y^2 +2z^2 -8xyz$$
Setting the partial derivatives to zero leads to:
$$x-yz=0 \\ y-4xz=0\\z-2xy=0$$
Substiting $z=2xy$ into the first two equations to get $y(1-8x^2)=0$ and $x(1-2y^2)=0$. Which have the solution $y=0,x=\frac{1}{\sqrt{8}}, x=-\frac{1}{\sqrt{8}}$ and $x=0, y=\frac{1}{\sqrt{2}},y=-\frac{1}{\sqrt{2}}$
Now how do I go about pairing my $x,y$ solutions?
 [/INST] You have two conditions:


*

*$y=0$ or $x=\frac{1}{\sqrt{8}}$ or $x=-\frac{1}{\sqrt{8}}$

*$x=0$ or $y=\frac{1}{\sqrt{2}}$ or $y=-\frac{1}{\sqrt{2}}$


Both of these must be true. That means that one of the conditions gives the value for $x$, while the other must give the value for $y$. For example, if $y=0$ then the first statement is true. Then $y\neq\frac{1}{\sqrt{2}}$ and $y\neq-\frac{1}{\sqrt{2}}$, so $x=0$ must be true to make the second statement true. 
So $(x,y)=(0,0)$ is possible and gives you a possible value for $z$. In a similiar way, $(x,y)=(\frac{1}{\sqrt{8}},\frac{1}{\sqrt{2}})$,$(x,y)=(\frac{1}{\sqrt{8}},-\frac{1}{\sqrt{2}})$, $(x,y)=(-\frac{1}{\sqrt{8}},\frac{1}{\sqrt{2}})$ and $(x,y)=(-\frac{1}{\sqrt{8}},-\frac{1}{\sqrt{2}})$ are possible. Each of them gives you a value for $xy$ and hence for $z$. 
Don't forget to check whether every solution indeed satisfies all equations in the system. You can do this by just calculating the given expressions with values for $x,y,z$. 
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