Finding extreme values of $f(x,y,z)=x^2+2y^2+3z^2$ on unit sphere $x^2+y^2+z^2=1$ 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. 

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 $f(0,0,\pm 1)=1+2=3\,$.
