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?
On
You have to be careful: a solution of this system is always of the form $(x,y,z)$ so you should try to do this more structurally.
After the substitution of $z=2xy$ in the first equation, you get $$x(1-2y^2) = 0$$ and from this, you have $x=0$ or $y=\tfrac{1}{\sqrt{2}}$ or $y=-\tfrac{1}{\sqrt{2}}$. This gives you three cases, check them all.
$\fbox{$x=0$}$
If $x=0$, then from the second equation you get $y=0$ and from the first you get $z=0$; this gives you a first stationary point: $\color{blue}{(0,0,0)}$.
Now check the other possibilities.
$\fbox{$y=\tfrac{1}{\sqrt{2}}$}$
(...)
$\fbox{$y=-\tfrac{1}{\sqrt{2}}$}$
(...)
You have two conditions:
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 $x,y,z$.