Lagrange multiplier - space probe i am stuck on this question which uses the Lagrange multiplier. I am trying to construct the equations using the partial derivatives but the $x$'s and $y$'s cancel. can anyone help?
A space probe in the shape of the ellipsoid
$x^2 + y^2 + 3z^2 = 3$
enters a planet's atmosphere and begins to heat up. The temperature on its surface is found
to be
$T(x, y, z) = x^2 + 2y^2 + 6z$:
Use the method of Lagrange multipliers to find the hottest points on the probe's surface.

Set $\Lambda \colon\mathbb R^4\to \mathbb R, (x,y,z,\lambda)\mapsto x^2+2y^2+6z+\lambda (x^2+y^2+3z^2-3)$.
Let $(x,y,z,\lambda)\in \mathbb R^4$.
The following holds:
$$\begin{cases} \Lambda _x(x,y,z,\lambda)&=2x+2\lambda x\\ \Lambda _y(x,y,z,\lambda)&=4y+2\lambda y\\ \Lambda_z(x,y,z, \lambda)&=6+6\lambda _z\\ \Lambda _\lambda(x,y,z,\lambda)&=x^2+y^2+3z^2-3.\end{cases}$$
Suppose $$\begin{cases} 0&=2x+2\lambda x\\ 0&=4y+2\lambda y\\ 0&=6+6\lambda z\\ 0&=x^2+y^2+3z^2-3,\end{cases}$$
then $$\begin{cases} 0&=(1+\lambda)x\\ 0&=(2+\lambda)y\\ \lambda&=-\dfrac1 z \land z\neq 0\neq \lambda \\ 0&=x^2+y^2+3z^2-3,\end{cases}$$
which implies $$\begin{cases} \lambda =-1\lor x=0\\ \lambda =-2 \lor y=0\\ \lambda=-\dfrac1 z \land z\neq 0\neq \lambda \\ x^2+y^2+3z^2-3=0.\end{cases}$$
$\boxed{\text{Case }\lambda =-1}$
It follows that $y=0$ and $z=1$. Thus $x^2+0+3-3=0$ and $x=0$, yielding the critical point $\color{blue}{(0,0,1)}$.
$\boxed{\text{Case }x=0}$


*

*$\boxed{\text{Sub case }\lambda =-2}$ It comes $z=\dfrac 1 2$ and $x^2+y^2-\dfrac 9 4=0$, that is $y=\pm\dfrac 3 2$, yielding the critical points $\color{blue}{\left(0, \dfrac 3 2, \dfrac 1 2\right)}$ and $\color{blue}{\left(0, -\dfrac 3 2, \dfrac 1 2\right)}$.

*$\boxed{\text{Sub case }y=0}$ It comes $0+0+3z~2-3=0$, that is, $z=\pm 1$, yielding the critical points $\color{blue}{(0,0,-1)}$ and $\color{blue}{(0,0,1)}$.


Now just check where it is hotter.
