The observation you made, about how these $\varepsilon$-$\delta$ proofs often come down to something multiplied to $|x - x_0|$, is a good one.
Let's consider rational functions $f = \frac{p}{q}$, where $p$ and $q$ are polynomials, and $q$ is not the zero polynomial. Why? Because this is very much a staple of $\varepsilon$-$\delta$ exercises. If you also add in radicals, e.g. square/cube roots, then you probably account for $99\%$ of $\varepsilon$-$\delta$ exercises involving specific functions that we can give to a student (including polynomials).
If we have such a function $f$, and some $x_0$ in the domain of $f$ (i.e. $q(x_0) \neq 0$), then we have
$$|f(x) - f(x_0)| = \left|\frac{p(x)}{q(x)} - \frac{p(x_0)}{q(x_0)}\right| = \frac{|p(x)q(x_0) - q(x)p(x_0)|}{|q(x)q(x_0)|}.$$
Note that $p(x)q(x_0) - q(x)p(x_0)$ is a polynomial with a root at $x_0$, which implies that it contains a factor of $x - x_0$. So, there is some polynomial $r(x)$ such that $p(x)q(x_0) - q(x)p(x_0) = (x - x_0)r(x)$, and so
$$|f(x) - f(x_0)| = \frac{|r(x)|}{|q(x)q(x_0)|}|x - x_0|.$$
Essentially, it's as you've observed: we get a big ugly thing, $\frac{|r(x)|}{|q(x)q(x_0)|}$, multiplied by $|x - x_0|$. This phenomenon is not unique to rational functions (it happens when dealing with radicals as well), but it does clarify what we need to do.
Since we need to make $|f(x) - f(x_0)|$ small, having only the ability to make $|x - x_0|$ small, we just need to control our big ugly thing. We need to ensure that, as we make $|x - x_0|$ very small, that $\frac{|r(x)|}{|q(x)q(x_0)|}$ does not become proportionally big.
This is where we observe the need the kind of inequality that you talk about in your question. We look to ensure that $\frac{|r(x)|}{|q(x)q(x_0)|} \le M$ for $x$ around $x_0$, and we then ensure that $\delta \le \frac{\varepsilon}{M}$.
The fact is, often these big ugly things can become very big for certain values. It's possible that, given $r$ is a polynomial, if $q$ has a lesser degree than $r$ (e.g. if $q$ is constant, and $f$ is a polynomial), this function might grow without bound as $x \to \pm \infty$. Also, if $q$ is not constant, the function will have asymptotes, so the function might grow without bound around finite values of $x$ as well. All of these must be avoided in order to ensure $|f(x) - f(x_0)|$ becomes small.
The good news is, none of this behaviour occurs around $x_0$. Since $q(x_0) \neq 0$, we know that $\frac{|r(x)|}{|q(x)q(x_0)|}$ has no asymptote at $x = x_0$. And, since $r(x)$ can only grow large for large values of $x$, it will only have controlled growth around $x = x_0$. To deal with both of these problems, we choose an upper limit on our $\delta$. We choose $\delta$ small enough that $[x_0 - \delta, x_0 + \delta]$ avoids any asymptotes (the interval is closed, because even having $x_0 \pm \delta$ be an asymptote will mean that the function maps points in $(x_0 - \delta, x_0 + \delta)$ to arbitrary large values). Making any restriction on $\delta$ will prevent $r(x)$ from growing too large as well.
As an example of this, with your function $f(x) = x + \frac{1}{x^3}$, we observe an asymptote at $x = 0$. So, inevitably, our choice of $\delta$ must be restricted to be less than $1$, the distance from $x_0 = 1$ and the asymptote $0$. Choosing $\delta = 1$ would still be a mistake, but anything less would be fine. The solution you've seen seems to involve choosing $\delta \le \frac{1}{2}$ (given the inequality involves $x_0 > 1 - \frac{1}{2} = \frac{1}{2}$, but choosing $\delta \le 0.999$ would be fine as well. It would just mean that our big ugly thing would grow to a larger maximum $M$ (given $x_0 - \delta$ could be as close as $0.001$ to the asymptote), meaning that our $\delta \le \frac{\varepsilon}{M}$ will have to become tighter (which isn't a problem: the limit is still proven).
Once you have suitably restricted your $\delta$, your big ugly thing should now be bounded. You should be able to find some $M$ such that $\frac{|r(x)|}{|q(x)q(x_0)|} \le M$, and the rest is easy.
So, hopefully, this explains why we would seek out inequalities like this, and what we expect from these inequalities. But, it doesn't address how we come up with these inequalities.
The "how" comes from a hodge-podge of methods, some of which are ad-hoc. You now have essentially an optimisation problem: you need to find the maximum/minimum of $\frac{r(x)}{q(x_0)q(x)}$, and find out which has the bigger absolute value. You don't even need to find exactly the maximum or minimum, just a number that you can prove is greater than their absolute values.
Even though you're proving continuity, which is weaker than differentiability, you could still employ the method of finding stationary points and evaluating them, as well as checking endpoints like $x_0 - \delta$ and $x_0 + \delta$. This will give you an upper bound, provided you can find roots. However, you will need to find an independent justification for your guess, because you can't assume differentiability to prove continuity! The method is just to give you sometihng to aim for: can you show that $\frac{|r(x)|}{|q(x)q(x_0)|}$ is actually less than this supposed maximum?
You can also do things like factor $r(x)$ and $q(x)$ into linear factors, and find the maximum value of each linear factor in $|r(x)|$, as well as the minimum value of each linear factor in $|q(x)|$, to get the greatest possible value of $\frac{|r(x)|}{|q(x_0)q(x)|}$.
There are other tricks as well that can be useful in certain situations (e.g. a deft use of triangle inequality can sometimes really help, or it could lead you up a blind path). But, in my experience, intelligently limiting the $\delta$ and facing up to your big ugly thing will usually give you a good idea of how best to limit it.
Anyway, I hope that helps.
