Using Lagrange multiplier to find all extrema

It is given that $$f(x,y,z)=x^2-y^2+z^2$$ and the constraint is $$g(x,y,z)=x^2+2y^2+3z^2-1=0$$ I started to take Lagrange multiplier approach, here is my work $$\left\{ \begin{array}{c} 2x-2xλ=0 \\ -2y-4yλ=0 \\ 2z-6zλ=0 \\ x^2+2y^2+3z^2-1=0 \end{array} \right.$$ Which gives $$\left\{ \begin{array}{c} x(1-λ)=0 \\ y(1+2λ)=0 \\ z(1-3λ)=0 \\ x^2+2y^2+3z^2-1=0 \end{array} \right.$$ Then, I have no idea how to plug equation (1) to (3) in the constrain since x, y and z equal to zero. If x, y and z equal to zero, then equation (4) will not work.  If x, y and z not equal to zero, then $$λ=1 \ or \ λ=-1/2 \ or \ λ=1/3 \$$ What's wrong in my approach, do I missed other constraints?

The system of equations that you obtained, that is $$\begin{cases} x(1-\lambda)=0 \\ y(1+2\lambda)=0 \\ z(1-3\lambda)=0 \\ x^2+2y^2+3z^2-1=0 \end{cases}$$ is correct. Now looking at $x(1-\lambda)=0$, you should say that if $x\not=0$ then $\lambda=1$ and $$\begin{cases} 3y=0 \\ -2z=0 \\ x^2+2y^2+3z^2-1=0 \end{cases} \implies y=z=0,\; x=\pm 1.$$ Can you take it from here and see what happens when $y\not=0$? What if $z\not=0$?
Note that the constraint $x^2+2y^2+3z^2-1=0$ implies that at least one of $x$, $y$, $z$ is different from zero.
now for each value of $\lambda$ you have to find a solution