This exercise asks you for map $$f:\mathbb{R}^3\rightarrow \mathbb{R}:\,f(x,y,z)=xyz^{2}$$
(a) find the critical points and the critical values,
(b) and tell for what values $c$ the set $f(x,y,z)=c$ is a regular surface.
Definition 2 on p. 58 of the book tells you that in order to enumerate all the critical points you need to examine the differential map $df$ which can be represented by a matrix as $$
(f_x,f_y,f_x)=(y z^2,x z^2, 2 x y z)$$
Now you can see that $df:\mathbb{R}^3\rightarrow \mathbb{R}$ is not surjective iff $(y z^2,x z^2, 2 x y z)=(0,0,0)$, so we can write down the set of critical points as
$$
\mathcal{K}=\{(x,y,x) \in \mathbb{R}^3|z=0 \text{ OR } (x=o \text{ AND } y=0)\}
$$
You can see immediately that the only critical value is $0=f(\mathcal{K})$.
In other words, the regularity fails in the standard way: "the derivative is zero."
This solves part (a).
As for part (b), you now need to argue that the set that you described in your question is not the graph of any function as Proposition 3 on p. 63 of the book suggests. For that it is enough to examine three cases, as shown in this Proposition: $z=f(x,y), y=g(x,z), x=h(y,z)$. It is easy to see that your set cannot be of any of these forms.
A confusion may occur due to an identification of the tangent spaces of $\mathbb{R}^n$ with $\mathbb{R}^n$ (I hope you can see where this happens in the above).
Proposition 2, p. 59, ibid., tells you that all other points will be regular.
Hopefully, this helps.
(Edited to add the correct demonstration of part (b)).
Best Answer
It is because in any neighborhood of a boundary point none of the orthogonal projections to the coordinate plane give a parametrization. This is a proposition in that section,and is often used to show that something is not a manifold.