Let $f:\mathbb{R}^3\to\mathbb{R}$ be given by $f(x,y,z)=z^2$. Prove that $0$ is not a regular point but $f^{-1}(\{0\})$ is a manifold.
I divided this in two parts:
$(1)\; 0$ is not a regular point
$0$ regular if $\nabla f(p)\neq 0\; \forall p:f(p)=0$, but here $f(p)=0$ iff $p=(x,y,0)$ with $x,y\in\mathbb{R}$ and $\nabla f=(0,0,2z)$ therefore $\nabla (p)=0\; \forall p:f(p)=0$.
$(2)$ $f^{-1}(\{0\})$ is a manifold.
I have to prove that is possible define a diffeomorphism to $\mathbb{R}^2$. I have $f^{-1}(\{0\})=\{(x,y,0):x,y\in\mathbb{R}\}$ and then I defined $g:f^{-1}(\{0\})\to\mathbb{R}^2$ by $g(x,y,0)=(x,y)$.
$g$ is bijective, because if $g(p)=g(q)$ then $(p_1,p_2)=(q_1,q_2)\implies p=q$ (using that the last component is zero $\forall p\in f^{-1}(\{0\})$.
Looking at the partial derivatives $\displaystyle\frac{\partial g_1}{\partial x}=\displaystyle\frac{\partial g_2}{\partial y}=1$ and $\displaystyle\frac{\partial g_1}{\partial y}=\displaystyle\frac{\partial g_2}{\partial x}=0$. And it possible to keep taking partial derivatives to the last ones being all of them continuous. Then $g$ defines a diffeomorphism.
Is this enough to conclude that $f^{-1}(\{0\})$ is a manifold?
Yes. The whole thing looks very well done to me, and very nicely written.
The key to part (2) is of course to realize that the set $f^{-1}(\{ 0 \}) = \{ (x, y, 0) \mid (x, y) \in \Bbb R^2 \}$, manifestly a manifold.
Hope this helps! Cheerio,
and as always,
Fiat Lux!!!