$$x^2+y^2+z^2-xz-yz-2x+2y-2=0. $$
So I have $$F(x,y,z)=x^2+y^2+z^2-xz-yz-2x+2y-2 \\ \frac{\partial F}{\partial z}=2z-x-y\neq 0 \\ \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}=-\frac{2x-z-2}{2z-x-y}=0 \\ \frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}=-\frac{2x-y+2}{2z-x-y}=0 \\ $$
So I basically have two linear equations with 3 unknowns, and I have no way, atleast right now, to know how to find the potential extremes, or stationary points. When I get $z=x+y$ and substitute this in the main equation, after simplifying I get $$(x-1)^2+(y+1)^2=4$$ a sphere. Can I do anything with this information??
I think you are looking at this the wrong way. You say that the function is given implicitly, so I would assume that $z=z(x,y)$. If we keep this is mind we can just do implicit differentiation. First taking the partial with respect to $x$ on both sides we get $$2x+2z\frac{\partial z}{\partial x}-z-x\frac{\partial z}{\partial x}-y\frac{\partial z}{\partial x}-2=0$$ Which gives $$\frac{\partial z}{\partial x}=\frac{z-2x+2}{2z-x-y}=0$$ Doing the same thing for $z_y$ we get $$\frac{\partial z}{\partial y}=\frac{z-2y-2}{2z-x-y}=0$$ Which has the solution $y=x-2$. Thus all critical points are of the form $(x,x-2)$.