As the title says, I would like to show that for any constant $k > 0$, there exists a $c>0$ such that the inequality $\frac{n}{e^{k n}}< \frac{c}{n^2}$ holds for any $n\in\mathbb{N}\setminus\{0\}$.
It is rather easy to see that such a $c$ exists when $k>1$, in which case $c=1$ would be sufficient. However, I am unsure how to start the problem for $k<1$.
On
$\dfrac{n}{e^{kn}} \lt \dfrac{c}{n^2} \iff n^3 \lt ce^{kn} $.
Consider the more general problem of showing that, for any $m, k > 0$ there is a $c = c(k, m)$ such that $n^m \lt ce^{kn} $ for all integer $n \ge 1$.
Taking the $k$-th root, this is $n^{m/k} \lt c^{1/k}e^{n} $, which is the same problem with $k=1$.
Therefore, for any $r > 0$, we want to find $d > 0$ such that $n^r \lt de^n$ for $n \ge 1$.
Here is a simple answer that is far from the best but still works:
Let $s = \lfloor r \rfloor +1$, so $n^r < n^s$ for all $n \ge 1$. Then, from the power series for $e^x$, $e^n \gt \dfrac{n^s}{s!}$.
Therefore, if $n^s \le d\dfrac{n^s}{s!}$ we are done. This is just $d \gt s!$.
Going back to the original statement, $s = \lfloor m/k \rfloor +1$ and we want $c^{1/k} > s!$ or $c > (s!)^k$.
Let $k > 0$.
Denote by $f : \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}$ be the function defined by $$f(x) = \frac{x^{3}}{e^{k x}}$$ Then $f$ is continuous on $\mathbb{R}_{\geq 0}$ and we have $$f(x) \underset{x \rightarrow +\infty}{\longrightarrow} 0$$ Therefore, $f$ is bounded above on $\mathbb{R}_{\geq 0}$.
Thus, for all integer $n \geq 1$, we have $$\frac{n}{e^{k n}} = \frac{f(n)}{n^{2}} < \frac{c}{n^{2}}$$ where $c = \sup\left\{f(x) : x \in \mathbb{R}_{\geq 0}\right\} +1$.