If $f:\mathbb{R}^2 \rightarrow \mathbb{R}^3$ is differentiable at $a$ such that the rank of $Df(a)$ is $2$, then the trace of $Df(a)$ is a plane tangent to the function $f$ at $a$.
The only examples I've been able to come up of a case where $Df(a)<2$ consist in the function $f$ tracing a curve, for example $$f:(x,y)\mapsto (x,2x,3x).$$
Is it possible to have a function $f:\mathbb{R}^2 \rightarrow \mathbb{R}^3$ such that $Df$ exists at $a$ with rank less than $2$ and there is a plane tangent to $f$ at $a$?
Put another way, does the fact that the rank of $Df(a)$ is less than $2$ necessarily imply that there is no tangent plant of $f$ at $a$?
Let $f(x,y) = (x^3,y^3,0)$. The image of $f$ is $\mathbb R^2 \times \{0\}$ which has itself as a tangent plane at all points. But $Df$ is $0$ at $(0,0)$.