# Finding extremes for function $f(x,y,z)$

I am preparing for an exam on Tuesday and I'm having problems with this excercise:

Decide, whether there are extremes for the function $f:\mathbb{R^3} \rightarrow \mathbb{R}, f(x,y,z) = xyz \hspace{2mm}$ on an elipsoid $3x^2 + 3y^2 + z^2 = 1 \hspace{2mm}$ If there are any, find them.

My solution so far:

$3x^2 + 3y^2 + z^2 -1 = 0$
$L(x,y,z, \lambda) = xyz - \lambda(3x^2 + 3y^2 + z^2 -1)$

Gradient:$\hspace{20mm}$Normal:
$f'_x = yz \hspace{21mm} N'_x =6x$
$f'_y = xz \hspace{21mm} N'_y =6y$
$f'_z = xy \hspace{21mm} N'_z =2z$

$(yz, xz, xy) = k(6x, 6y, 2z)$

$yz = 6kx$
$xz = 6ky$
$xy = 2kz$
$3x^2 + 3y^2 + z^2 = 1$

And I have a really hard time getting the points out of these four equations - I always run into a dead end. Can someone please give me a hint? Also I'm sorry for the formatting, but I haven't used LateX in a while.

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Consider multiplying your first equation by $x$, the next by $y$, and the third by $z$. You'll get
$xyz=6kx^2$, $xyz=6ky^2$, and $xyz=2kz^2$. Now dividing each by $2k$ you get that $3x^2=3y^2=z^2$. At this point you can plug them into the constraint equation and find the answer.
$3x^2+3y^2+z^2=1$ becomes $z^2+z^2+z^2=1$ so that $z^2=1/3$ and so $z=\pm 1/\sqrt{3}$. Next $3x^2=z^2=1/3$ etc.