Consider a curve created by the intersection between the ellipsoid $x^2 + 2y^2 + z^2 = 1$ and the plane $x+y=1$.
I want to find the points on that curve for whch it's greatest and smallest the distance between those point and the origin.
Here is what I tried to do, following Robert Z comment. I started with the function $f(x, y, z) = x^2+y^2+z^2$ subject to the two constraints and hence imposing Lagrange multipliers conditions. Building the Lagrangian
$$L(x, y, z, t, \lambda) = f(x, y, z) - \lambda g(x, y, z) - \mu h(x, y)$$
$$L(x, y, z, \lambda, \mu) = x^2+y^2+z^2 - \lambda(x^2 + 2y^2 + z^2 - 1) + \mu(x+y-1)$$
Thence
$$\begin{cases} 2x = 2\lambda x + \mu \\ 2y = 4\lambda y + \mu \\ 2z = 2\lambda z \\ x^2+2y^2+z^2 = 1\\ x+y = 1 \end{cases} $$
This gives immediately $\lambda = 1$ and hence $\mu = -2y$ by using that value for $\lambda$ in the second equation.
Then I tried to solve the first one for $x$, expliciting $y = 1-x$ from the last.
This gives $x = 1$ and hence $y = 0$.
In the ellipsoid constraint, finding $z$ returns to be $z = 0$
At this point I only got one point: $$P = (1, 0, 0)$$
And from this I don't know what exactly to do, for I expectd to obtain two points, a max and a min for which the distance from the origin is max and min...
Your work is not so far from the conclusion.
The equation $2z = 2\lambda z$, that is $2z(1-\lambda)=0$, gives TWO cases: $\lambda=1$ (in your work) OR $z=0$. Putting $z=0$ in the system we find
$$\begin{cases} x^2+2y^2 = 1\\ x+y = 1 \end{cases} \implies P=(1,0,0) \lor Q=(1/3,2/3,0).$$ So we have TWO constrained stationary points $P$ and $Q$. Now the given curve is compact (it is the intersection of a compact ellipsoid and a closed plane) and regular. Hence the maximum and minimum points exist and they are stationary points. It remains to compare the distances of $P$ and $Q$ from the origin.