How to know when you can focus only on a specific part of an expression with an even power?

In your solution, when you perform the step $$ x^4 = 3x \iff \frac{x^4}{x} = x^3 = 3, $$ you are using the law of multiplicative cancelation, which states

Theorem: If $a$, $b$, and $c$ are any real numbers with $c \ne 0$, then $$ a = b \iff ac = bc. $$

In other words, we can cancel common factors from both sides of an equation (that is, we can "divide" both sides of an equation by something), assuming that this factor is nonzero. In your computation, you are applying this theorem. In order for this to be a valid step, the hypotheses of the theorem much hold, hence you need it to be true that $x \ne 0$. Therefore you are implicitly assuming that $x \ne 0$.

Since you ask about "guidelines" for working problems like this, it might be worthwhile to get extremely pedantic, and very carefully reason about every step. For example, your solution could be expanded to something akin to the following:

Let $$ x = \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}}, $$ assuming that $x$ exists.^[1] Then $$ x^4 = \left( \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}} \right)^4 = 3 \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}} = 3x. $$ Since $x > 0$,^[2] the law of multiplicative cancelation gives $$ x^4 = 3x \iff x^3 = 3. $$ The function $f(x) = x^3$ is invertible on $\mathbb{R}$, and has inverse $f^{-1}(x) = \sqrt[3]{x}$. Thus $$ x^3 = 3 \iff x = \sqrt[3]{3}. $$ Therefore $x = \sqrt[3]{3}$.

This kind of solution is overly pedantic, but each and every step of the computation is justified by appealing to some mathematical principle. Often we skip writing these justifications out explicitly, as we can eventually start assuming that we all speak a common language and are familiar with the same rules. However, it is sometimes helpful to give all the details, if for no other reason than to confirms one's one understanding.

Footnotes:

There are a couple steps in the computation which should probably be justified, but the justifications require more advanced mathematics (calculus, at least). You can skip the footnotes and be okay, but they are here for completeness.

[1] One can actually write down funny expressions like this which don't "converge" to real numbers. However, rigorously checking that such an expression gives an actual number requires a firm grasp of the concept of a limit, which is not generally introduced until calculus. So we just have to assume that $x$ really exists.

[2] It seems obvious that $x \ne 0$, but this, also, has the potential to be a little delicate. Basically, $x$ is the limit of iterative application of the map $t \mapsto \sqrt[4]{3t}$. This map has two fixed points (one at zero, and one somewhere else—the second fixed point is the thing that we are trying to find). However, the fixed point at zero is "unstable", in the sense that repeated application of the map won't ever give a sequence converging to zero, unless the very first thing we plug in is zero.

For $ab=0$, either $a=0$, or $b=0$ or both $a=b=0$ (if possible). Here, you see, $x>0$, as $x$ is an even root (so $x$ is definitely not zero). This implies the other term in the product zero. So $x^3-3=0$. Whenever you have equation of the form $a_1a_2a_3...a_n=0$, always make cases.