I cannot suggest any particular book, but I can find the region $R$ for the specific question, assuming I interpreted it correctly.
Most introductory books on calculus and analysis should contain sufficient tools to come up with the following solution.

This sequence is of a very special form.
I would not call it a multivariable sequence, but a usual single variable sequence with two real parameters.
(I assume that $x$ and $y$ are real numbers in your question.)

If we denote $f(x,y)=e(\sqrt y-\sqrt x)^2/x$, the question is when does $\lim_{n\to\infty}|f(x,y)^n|=0$.
This is true if $|f(x,y)|<1$ and false if $|f(x,y)|\geq1$.
The case $|f(x,y)|=1$ is more delicate if you want to find the existence and the exact value of the limit.

To make sense of both square roots and dividing by $x$, we need to assume $x>0$ and $y\geq0$.
Then $f(x,y)\geq0$.
Now
$$
\begin{split}
&
|f(x,y)|<1
\\\iff&
f(x,y)<1
\\\iff&
e(\sqrt y-\sqrt x)^2/x<1
\\\iff&
e(\sqrt{y/x}-1)^2<1
\\\iff&
-1/\sqrt e<\sqrt{y/x}-1<1/\sqrt e
\\\iff&
(1-1/\sqrt e)^2<y/x<(1+1/\sqrt e)^2.
\end{split}
$$
Similarly, $|f(x,y)|>1$ is equivalent with $y/x\notin[(1-1/\sqrt e)^2,(1+1/\sqrt e)^2]$.
The borderline case is $y/x=(1\pm1/\sqrt e)^2$.
In the borderline case we obtain simply $f(x,y)=1$, so we have
$$
\lim_{n \to \infty} \; \;\left| e^n\frac{(\sqrt{y}-\sqrt{x})^{2n}}{x^n} \right|
=
\begin{cases}
0, & (1-1/\sqrt e)^2<y/x<(1+1/\sqrt e)^2\\
1, & y/x=(1\pm1/\sqrt e)^2\\
\infty, & y/x\notin[(1-1/\sqrt e)^2,(1+1/\sqrt e)^2]
\end{cases}
$$
under the assumption that $x>0$ and $y\geq0$.
The limit is always infinite if $y=0$.