Let $M \subseteq \mathbb{R}^3$ and $$M = \left\{ (x,y,z) \in \mathbb{R}^3:f(x,y,z)=0 \right\}$$ with $$f(x,y,z):=3z+3xy^2-x^3$$ Show that $M$ is a manifold and determine its dimension.
$f:\mathbb{R}^3 \to \mathbb{R}$ is continuously differentiable with $\nabla f(x,y,z)=(3y^2-3x^2,6xy,3)$
How can I argue that $\nabla f(x,y,z)$ has full rank?
As $f: \mathbb{R}^3 \to \mathbb{R}$, the dimension of $M$ should be $3-1=2$.
The definition we use:
Let $O \subseteq \mathbb{R}^p$ open and $f:O \to \mathbb{R}^{p-d}$ continuously differentiable with $0<d<p$. If $\nabla f(x) \in \mathbb{R}^{(p-d)\times p}$ has full rank for all $x \in O$ with $f(x)=0$, which means rank $p-d$, then $M:=\{x \in O:f(x)=0\}$ is a $d$-dimensional manifold if $$
You got it right! The only remaining thing is that the vector $\nabla f$ is full-rank. But a vector is full-rank precisely iff it is nonzero (in this context, full means rank-1, since vectors are, in a sense, $d\times 1$-matrices), and
$$\nabla f(x,y,z)=(3y^2-3x^2,6xy,3)$$
clearly cannot be zero (the last coordinate is always $3$).