Edited in response to @Randall's question, although I'm still not sure if this question makes sense.
I am trying to find out whether there exists a simple way to tell if a function from $\quad{\mathbb R}^n\to{\mathbb R}^n$ is strictly increasing. For example, with functions from $\quad{\mathbb R}\to{\mathbb R}$, I would take the first derivative and try to show that it is positive on the interval I'm interested in.
In higher dimensions, the function would have to be from $\quad{\mathbb R}^n\to{\mathbb R}^n$ so that the input and output vectors would be of the same dimensions. Then (as I understand it) the function $f : \mathbb{R}^n \to \mathbb{R}^n$ is strictly increasing if for all $\mathbf{x_1}, \mathbf{x_2} \in \mathbb{R}^n, \mathbf{x_1} \gt\gt \mathbf{x_2}$, we have $ \mathbf{y_1}=f(\mathbf{x_1}) \gt\gt \mathbf{y_2}=f(\mathbf{x_2})$, where $\mathbf{x_1} \gt\gt \mathbf{x_2}$ means that each component of $ \mathbf{x_1}$ is greater than than the corresponding component of $\mathbf{x_2}$.
Is there a test like looking at the first derivative, but in higher dimensions? Thanks in advance for the help.
If I understand you question correctly, my answer should provide a sufficient condition. If $f:\Bbb R^n\to\Bbb R^n$ is differentiable, then it is increasing if $$ \frac {\partial f_j(\xi)}{\partial x_i} \ge 0 \quad\text{for all $i,j=1,2,\dots,n$} $$ and all vectors $\xi\in \Bbb R^n$. You can replace the $\ge$ sign with $>$ if you want your function to be strictly increasing.
The proof that this works is as follows: Suppose that $\mathbf x<<\mathbf y$ then each component of $\mathbf v=\mathbf y -\mathbf x$ is greater than zero. By the fundamental theorem of calculus, $$\begin{align} f(\mathbf y) &= f(\mathbf x) + \left(\int_0^1 Df(\mathbf x + t\mathbf v) \,dt\right)\mathbf v. \end{align}$$ The above means that for each $j$ we have $$\begin{align} f_j(\mathbf y) - f_j(\mathbf x) = \sum_{i=1}^n \int_0^1\frac {\partial f(\mathbf x + t\mathbf v)}{\partial x_i}\, v_i \,dt \ge 0 \end{align}$$ by our assumption on the derivatives of $f$.