Let $f: \mathbb{R}^1 \mapsto \mathbb{R}^n$ be a smooth vector-valued function. Consider the $n \times n$ matrix $A(x)$ obtained from a vector $x \in \mathbb{R}^n$ by appropriately stacking $[f(x_1),\dots, f(x_n)]$.
The question is: what are easily verifiable sufficient conditions which guarantee that the matrix $A$ is generically of full rank, that is, $A(x)$ is full rank except possibly for points $x$ in sets of measure zero?
To fix ideas, obviously $f$ cannot be linear, otherwise there is linear dependence in $A$.
Since that was a bit too long for a comment, I wrote it as an answer. I think there is no general answer for this general question. Still that question is really interesting. So if I miss something, please let me know in the comments.
To state the obvious: $f$ can't be linear. Since otherwise, assuming w.l.o.g. $x_2\not=0$, it is $f(x_1)- \frac{x_1}{x_2}f(x_2)=0$. Therefore $f(x_1)$ and $f(x_2)$ are linear dependant, and thus $R$ has not full rank.
Even if $f$ is non-linear, there are simple examples where $R$ has not full rank. Let $f:ℝ→ℝ^2$, with $x↦(x^2,x^4)$. (It is not necessary to consider $ℝ^4$.)
Then it is $f(x) = f(-x)$. Choosing $x_1=x$ and $x_2 = -x$ it is $x_1\not=x_2$ and $R$ is not of full rank.