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I defined $G(x)=x^2+y^2+z^2-1$, such that the gradient of G $\nabla G=(2x, 2y, 2z)$, so that $\nabla f=(2x, 4y, 6z)=\lambda (2x,2y,2z)$.

The conclusion I drew was that the only possible value was either $\lambda=0$, or y and z were both 0 but x can be anything. This seems incorrect to me, but I'm not sure how to proceed.

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    You don't even need to use Lagrange multipliers. If $x^2 + y^2 + z^2 = 1$ then $f(x,y,z) = 3 - 2x^2 - y^2$. Now you just need to find the extreme values of this function on the domain $x^2 + y^2 \le 1$. – 2017-01-20

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You can write the Lagrangian as:

$$\mathcal L(x,y,z,\lambda)=x^2+2y^2+3z^2-\lambda(x^2+y^2+z^2-1)$$

So we can get the gradient and putting it equal zero:

\begin{cases}2x-2x\lambda=0\\4y-2y\lambda=0\\6z-2z\lambda=0\\x^2+y^2+z^2=1 \end{cases}\begin{cases}2x(1-\lambda)=0\\ 2y(2-\lambda)=0\\2z(3-\lambda)=0\\x^2+y^2+z^2=1\end{cases}

and we have these solutions: $(\pm1,0,0,1)$, $(0,\pm1,0,2)$ and $(0,0,\pm1,3)$.

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Without calculus: $f(x,y,z)=x^2+2y^2+3z^2 = 2(x^2+y^2+z^2) + z^2-x^2= z^2-x^2+2\,$.

Since $0 \le x^2,z^2 \le 1$ it follows that $-1\le z^2-x^2 \le 1$. The extrema are actually attained at $f(\pm 1, 0, 0) = -1+2 = 1$ and $f(0,0,\pm 1)=1+2=3\,$.

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Well, you can generalize to two of $x,y,z$ being $0$, and then choose the Lagrange multiplier as appropriate. But remember the constraint, if two of them are $0$, the last has to be $1$ on the unit sphere.

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    Right, that was silly of me. So then would the correct conclusion be that the extreme values are at $(1,0,0), (0,1,0),$ and $(0,0,1)$ since $\lambda$ is then arbitrary? – 2017-01-20
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    Yeah, though tack on the $-1$ possibilities as well, I did miss that. – 2017-01-20