Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ be a vector-valued function of $n$ variables, and $x_0\in $ Dom$f $. Is the following proposition true?
$$ \lim_{x\rightarrow x_0}\frac{f(x)}{||x-x_0||}=\underbrace{(0,..,0)}_{m\space times}\in \mathbb{R^m} \iff \lim_{x\rightarrow x_0}\frac{||f(x)||}{||x-x_0||}=0\in \mathbb{R}$$
If yes, how do you prove it? If not, can you show a counterexample?
First, dividing by $\|x - x_{0}\|$ is a red herring (i.e., an unnecessary complication).
Second, your proposition is true.
To show $f(x) \to (0, \dots, 0)$ implies $\|f(x)\| \to 0$, observe that the norm function is continuous (perhaps by expressing the norm as a composition of continuous functions), and the zero vector has norm $0$.
For the converse, note that for each $i = 1, \dots, m$, we have $|y_{i}| \leq \|y\|$. If $\|f(x)\| \to 0$, it follows at once that each component of $f(x)$ goes to $0$.