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.
