Maximum & minimum values of multivariable function I am to check for max & min values for the given function $$f(x, y, z) = xy^2z^3$$ which is defined on $$M = \left\{x, y, z > 0, x+2y+3z=6\right\}$$
So.. what I did is:
$F(x, y, z, \gamma) = xy^2z^3 - \gamma(x+2y+3z-6)$ therefore
$\begin{cases} \frac{\partial F}{\partial x} = y^2z^3 - \gamma = 0\\ \frac{\partial F}{\partial y}  = 2xyz^3-2\gamma = 0\\ \frac{\partial F}{\partial z} = 3xy^2z^2 -3\gamma = 0\\ \frac{\partial F}{\partial \gamma} = -x-2y-3z+6 = 0\\ \end{cases}$
After the calculation I get the possible point is $P = (1, 1, 1)$.
So I count the 2nd derivatives : 
$$\left[\begin{array}{ccc}
\frac{\partial^2 f}{\partial x^2} = 0 &\frac{\partial^2 f}{\partial x\partial y} = 2yz^3 &\frac{\partial^2 f}{\partial x\partial z} = 3y^2z^2 \\
\frac{\partial^2 f}{\partial y \partial x} = 2yz^3 &\frac{\partial^2 f}{\partial y^2} = 2xz^3 &\frac{\partial^2 f}{\partial y\partial z} = 6xyz^2 \\
\frac{\partial^2 f}{\partial z \partial x} = 3y^2z^2 &\frac{\partial^2 f}{\partial z \partial y} = 6xyz^2 &\frac{\partial^2 f}{\partial z^2} = 6xy^2z 
\end{array}\right]$$
So with our point P it looks like :
$$\left[\begin{array}{ccc}
0&2&3\\
2&2&6\\
3&6&6
\end{array}\right]$$ 
And here's where I got lost. $\det_{1} = 0$ therefore we have no idea whether there is an extremum or not, but Wolfram says there's a local maximum in $(1, 1, 1)$. How should I obtain it?
 A: Upon poking around a bit more, it looks like having H = 0 (what you call $ \ \det_1 \ $ ) is one of those situations where the standard "second derivative test" for three variables can't be applied.  We can fall back on substituting the constraint into the function to produce  
$$h(y,z) \ = \ (6 - 2y - 3z)  \cdot  y^2  \cdot  z^3 \ = \ 6y^2z^3 \ - \ 2y^3z^3 \ - \ 3y^2z^4 , $$
for which the discriminant is
[EDIT]
$$D \ |_{(1,1)} \ = \ [ \ h_{yy} \ h_{zz} \ - \ (h_{yz})^2 \ ] \ |_{(1,1)} $$
$$= \ [ \ (12z^3 - 12yz^3 - 6z^4)\cdot (36y^2z \ - 12y^3z \ - 36y^2z^2) $$
$$ - \ (36yz^2 - 18y^2z^2 - 24yz^3)^2 \ ]  \ |_{(1,1)} $$
$$ = \ (-6)(-12) \ - \ (-6)^2 \ = \ 36 \ > \ 0  \ . $$
Since $ \ h_{yy} \ |_{(1,1)}  =  -6 \ < 0  \ , $ this identifies this critical point as a local maximum in $ \ x \ $ under the constraint.  
The constraint surface is a bit peculiar, as this maximum sits atop a rather narrow "ridge" near a very deep "drop-off" (if one may discuss the "terrain" of the function).  This may explain the difficulty in applying the usual test.  Below are two views of $ \ x = h(y,z) \ $ .

[I am making this revision as I found an error in a coefficient of one of the second derivatives since my original posting.  There was also a typo in the Hessian matrix in OP's post, but this proved to be immaterial to the overall problem.]
A: You can use Lagrange multiplier method and here is the answer provided by CAS
$$ [x=6-3\,z,y=0,z=z,\lambda_{{1}}=0,x{y}^{2}{z}^{3}=0]$$
$$[x=-2\,y+6,y=y,z=0,\lambda_{{1}}=0,x{y}^{2}{z}^{3}=0]$$
$$[x=1,y=1,z=1,\lambda_{{1}}=1,x{y}^{2}{z}^{3}=1].$$
A: Rewrite your constraint as $x+y+y+z+z+z=6$ and apply the am/gm inequality, yielding $xy^2z^3\le 1$ ,with equality only when $x=y=z=1$ . The critical point is thus an absolute maximum over the range.
