Suppose we are to find the points nearest to the origin on the curve of intersection of the two surfaces $g^{-1}_{1}\{ 0 \}$ and $g_{2}^{-1}\{ 0 \}$, where $g_{1}: (x, y, z) \mapsto x^{2} - xy + y^{2} - z^{2} - 1$ and $g_{2}: (x, y, z) \mapsto x^{2} + y^{2} - 1.$ Since on $g_{1}^{-1}\{ 0 \} \cap g_{2}^{-1}\{ 0 \}$ we have $xy + z^{2} = 0,$ we may define $g: (x, y, z) \mapsto xy + z^{2},$ so that $g^{-1}\{ 0 \} = g_{1}^{-1}\{ 0 \} \cap g_{2}^{-1}\{ 0 \}.$ Then we are to minimize the function $f: (x, y, z) \mapsto x^{2} + y^{2} + z^{2}$ on $g^{-1}\{ 0 \}.$
If $(x, y, z) \in g^{-1}\{ 0 \}$ such that $f(x, y, z)$ is a local extremum and $\nabla g(x, y, z) \neq 0$, then there is a $\lambda \in \mathbb{R}$ such that $$\lambda \nabla f(x, y, z) = (2\lambda x, 2\lambda y, 2\lambda z) = \nabla g(x, y, z) = (y, x, 2z)$$
However, it seems to me that the vector equation above in combination with the constraint $g(x, y, z) = 0$ lead to $x = y = z = 0.$ Where did I commit an error?
I didn't really see any error in your computation. However you shouldn't drop one constraint. Instead you should keep the two simpler constraints
$$g_2(x,y,z)=x^2+y^2-1\\ g(x,y,z)=xy+z^2$$
Use two parameters for Lagrange multiplier:
$$L=f(x,y,z)+\lambda g_2(x,y,z)+\mu g(x,y,z)$$
Then set up:
$$\nabla f+\lambda \nabla g_2 +\nabla g=0$$
together with the two constraints.
Try to continue from there. There are several cases to consider. You will get nontrivial solutions.
To answer your question about the geometry: You can try letting $x,y,z$ be constants and visualize the cross-sections. They seem like parabola and hyperbola, and shrinking toward the origin. Now including the cylinder $x^2+y^2=1$, you'll get a closed curve (probably two). Notice that your objective function has level surfaces as spheres. Shrinking or enlarging them to touch the curve would give some maxima and\or minima.