An unusual method for multivariable extrema I'm going over some problems from my textbook and the problem states simply states to find the extrema for the function
$$f(x, y)=x^2y^3(6-x-y)$$
I'll skip over the initial part because it's simple enough and I do understand it, the partial derivatives are:
$$f^{'}_{x} = xy^3(-3x+2y+12)$$
$$f^{'}_{y} = x^2y^2(-3x-4y+18)$$
Solving the system gives us  one critical point $(2, 3)$ and two families of critical points $M_t(0, t)$ and $N_t(t, 0)$ for any $t \in \mathbb{R}$. The second partial derivative test will work for the point $(2, 3)$ showing that it is a local maximum. However, it won't work for the other two. So I flipped over to the solutions in the textbook and they state the following:

If we use the fact that $f(M_t) = f(N_t) = 0$ we notice that it's
  sufficient to check the sign of the function in the surroundings of
  these points. For example, if the function turns out to be positive
  around $M_{t_0}$ for some $t_0$ that would mean that we have a local
  maximum (and other way around).

Thinking about this in 2D I realized it does make sense, since the value of the point is exactly at $0$. Moving on:

The sign of the function $f(x, y) = x^2y^3(6-x-y)$ depends on the sign
  of the expression $6-x-y$. Now it's clear that for $t>6$ and $t<0$
$M_t$ are points of local maxima, where for $0<t<6$ they're the points
  of local minima. For $t = 0$ and $t = 6$ there are no extrema since
  for no matter how small surrounding of these points f takes both
  positive and negative values. We similarly prove that no point from
  the $N_t$ family is an extremum.

The very first sentence is confusing already. Doesn't the sign of the function also depend on the $x^2y^3$ term (well really only on $y^3$ term but still)? Everything after that is even more confusing (what they say 'it's clear from here' doesn't seem to be clear to me at all).
So does anyone have a better/more detailed explanation?
Thanks.
 A: The rule must be:

For example, if the function turns out to be positive around $M_{t_0}$ for some $t_0$ that would mean that we have a local $\color{red}{minimum}$ (and other way around).

The sign of the function does depend on $y^3$. 
Indeed, given $f(x, y) = x^2y^3(6-x-y)$, at around $(0,t)$:
$$t>6 \Rightarrow t=6+\delta, 0<\delta<\infty\Rightarrow \\
f(0\pm\epsilon,(6+\delta)\pm \sigma)\stackrel{\delta>\sigma}=\underbrace{(0\pm\epsilon)^2}_{>0}\underbrace{((6+\delta)\pm \sigma)^3}_{>0}\underbrace{(6-(0\pm \epsilon)-((6+\delta)\pm \sigma))}_{<0}<0 \Rightarrow \\
f(0,t>6) \ \text{are the points of local maxima.}\\
t<0\Rightarrow t=0-\delta\Rightarrow \\
f(0\pm\epsilon,(0-\delta)\pm \sigma)\stackrel{\delta>\sigma}=\underbrace{(0\pm\epsilon)^2}_{>0}\underbrace{((0-\delta)\pm \sigma)^3}_{<0}\underbrace{(6-(0\pm \epsilon)-((0-\delta)\pm \sigma))}_{>0}<0 \Rightarrow \\
f(0,t<0) \ \text{are the points of local maxima.}\\
0<t<6\Rightarrow t=0+\delta=6-\phi\Rightarrow \\
f(0\pm\epsilon,(0+\delta \ \text{or} \ 6-\phi)\pm \sigma)\stackrel{\delta>\sigma \ \text{or} \ \phi>\delta}=\\
=\underbrace{(0\pm\epsilon)^2}_{>0}\underbrace{((0+\delta \ \text{or} \ 6-\phi)\pm \sigma)^3}_{>0}\underbrace{(6-(0\pm \epsilon)-((0+\delta \ \text{or} \ 6-\phi)\pm \sigma))}_{>0}>0 \Rightarrow \\
f(0,0<t<6) \ \text{are the points of local minima.}$$
