I am trying to understand example 9.6 from Tu text on manifolds. The problem is trying to expresss $S^2 = \{(x,y,z) \in \mathbb{R}^3 | x^2 + y^2 + z^2 = 1\}$ as a zero set. Thus, we define the equation $f(x,y,z) = x^2 + y^2+z^2-1=0$ and then clearly, $S^2 = f^{-1}(\{0\})$. This is the part I don't quite get, the text says: since $\frac{\partial f}{\partial x} = 2x$, $\frac{\partial f}{\partial y} = 2y$, $\frac{\partial f}{\partial z} = 2z$, the only critical point of $f$ is $(0,0,0)$.
I know that a critical point $p$ is a point where the differential of a map fails to be surjective. But I really see no connection between this definition and the quick conclusion that TU arrived at. The way I am trying to understand this is as follows: since $f:\mathbb{R}^3 \rightarrow \mathbb{R}$, then the differential of $f$, will also be a map from a $3$ dimensional vector space to a 1 dimensional vector space (well the tangent spaces). Thus, the only way this map has rank less than 1 (and so would not be a surjective map because the map doesn't have maximal rank) happens when we consider the point $(0,0,0)$. Thus, the point $(0,0,0)$ is the only critical point.
Is this reasoning correct for this type of problems or how can it be improved/made correct? Thanks for your help!
At any point $p$, the differential $df_p = (2x, 2y, 2z)$ which is a linear transformation $df_p: \mathbb R^3 \to \mathbb R$. It should be clear the only case it is not surgective, i.e., $\text{range}(df_p) = \{0\}$, is $x=y=z=0$.