Find local extremum and saddle points of the function $f(x, y) = x^2y^3(6 − x − y)$ $\nabla \left(f\right) = \begin{bmatrix}y^3x\left(12-3x-2y\right)\\ x^2y^2\left(18-3x-4y\right)\end{bmatrix}$
and in $(0,a), (a,0), (2,3)$ points $\nabla \left(f\right) = 0$ for all real a.
Hessian for $f$ is $\begin{bmatrix}2y^3(6-3x-y)\ \ xy^2(36-9x-8y)\\ xy^2(36-9x-8y) \ \ x^2y(36-6x-12y)\end{bmatrix}$
$(2,3)$ is local minimum but for $(0,a)$ and $(a,0)$ Hessian has $0$ eigenvalues.
https://www.wolframalpha.com/input/?i=saddle+points+of+x%5E2y%5E3%286+%E2%88%92+x+%E2%88%92+y%29&fbclid=IwAR3fEvSQ7zjGIgwCKWoPa3ZdazCPIhlDdbLiUNbH0orZUqt2p__qiRA1fE0
This says that $(2,0)$ is saddle point but how can I find it.And what can I say about $(0,a)$ and $(a,0)$ points?
 A: Check this graph out. The purple region indicates where $f(x,y)>0$ while the red region indicates where $f(x,y)<0$. From the graphic it's clear that any point on the $y-$axis different from $(0,0)$ and $(0,6)$ is either a local max or a local min, while any point on the $x-$axis is a saddle point. The point $(0,a)$ is a local max if $a\in(-\infty,0)\cup(6,\infty)$ while the point $(0,a)$ is a local min if $a\in(0,6)$. The points $(0,6)$ and $(0,0)$ are saddle point.
A: In the cases where the Hessian is $0$ you can try direct inspection, though the identification isn't always simple. Remember $p$ is a saddle point of $f$ iff $p$ is a critical point but not a local extremum, that is, every neighborhood of $p$ takes values both lower and greater than $f(p)$.
Here $f(a,0) = f(0,a) = 0$, so to have a saddle point $f$ must change signs around the critical point. But since $f$ is already conveniently factored, it is clear the sign will change when either $y = 0$ or $x+y = 6$, so the saddle points will end up being $(a,0)$ and $(0,6)$. This can be verified more rigorously by taking an open ball around each of those points and then identifying directions that will result on $f$ taking opposing signs inside the ball.
For the remaining critical points, just look at the sign of the expression
$$
y < 0 \implies f(x,y) = x^2y^3(6-x-y)<0 \text{ around } x=0\implies \text{local minima at } (0,a), a<0\\
0 < y < 6 \implies f(x,y) = x^2y^3(6-x-y)>0 \text{ around } x=0\implies \text{local maxima at } (0,a), 0<a<6\\
y > 6 \implies f(x,y) = x^2y^3(6-x-y)<0 \text{ around } x=0\implies \text{local maxima at } (0,a), a>6\\
$$
