Question: $f:\mathbb{R}\to\mathbb{R}$ satisfies $f(3x)=3f(x)-4f(x)^3, \forall x\in\mathbb{R}$ and is continuous at $x=0$. Show that $|f(x)|\le 1, \forall x.$
Solution: For the sake of contradiction, let us assume that $|f(x)|>1$ for some $x\in\mathbb{R}$. Thus, $|f(x)|=\left|3f\left(\frac{x}{3}\right)-4f\left(\frac{x}{3}\right)^3\right|>1.$
Hence, either $3f\left(\frac{x}{3}\right)-4f\left(\frac{x}{3}\right)^3>1$ or $3f\left(\frac{x}{3}\right)-4f\left(\frac{x}{3}\right)^3<-1.$
Observe that from here we can conclude that either $f\left(\frac{x}{3}\right)>1$ or $f\left(\frac{x}{3}\right)<-1$, i.e, $\left|f\left(\frac{x}{3}\right)\right|>1.$
Now since $x$ was arbitrarily chosen, we can conclude that $\left|f\left(\frac{x}{3^n}\right)\right|>1, \forall n\in\mathbb{N}.$
Now, since, $f$ is continuous at $0$, by the definition of limit after Heine and definition of continuity, we can conclude that for any sequence $\{a_n\}_{n\ge 1}$, if $\lim_{n\to\infty}a_n=0,$ then $\lim_{n\to\infty}f(a_n)=f(0)$. Now, since $\lim_{n\to\infty}\frac{x}{3^n}=0\implies \lim_{n\to\infty}f\left(\frac{x}{3^n}\right)=f(0).$
Also since we have already shown that $$\left|f\left(\frac{x}{3^n}\right)\right|>1, \forall n\in\mathbb{N}\\\implies \lim_{n\to\infty}\left|f\left(\frac{x}{3^n}\right)\right|=\left|\lim_{n\to\infty}f\left(\frac{x}{3^n}\right)\right|=|f(0)|\ge 1.$$
Now substituting $x=0$ in the equality $f(3x)=3f(x)-4f(x)^3,$ we have $$f(0)=3f(0)-4f(0)^3\\\implies f(0)(2f(0)^2-1)=0\\\implies f(0)=0\text{ or }f(0)=\frac{1}{\sqrt{2}}\text{ or }f(0)=-\frac{1}{\sqrt 2}.$$ Thus, in any case, we have $|f(0)|<1$. Hence, we obtain a clear contradiction.
Hence, we can conclude that $|f(x)|\le 1, \forall x\in\mathbb{R}.$
Is this solution correct and rigorous enough? And, is there any other way to solve the same?
Your proof is fine. One essential step (which you might want to emphasize) is that $$ | 3t - 4t^3 | > 1 \implies |t| > 1 $$ for all real numbers $t$, and that is because the polynomial $p(t) = 3t - 4t^3$ maps the interval $[-1, 1]$ into itself.
You can also argue the other way around: $|f(0)| < 1$ and the continuity at $x=0$ implies that $|f(x)| \le 1$ on some interval $(-\epsilon, \epsilon)$, and then repeated application of $$ |f(3x)| = |p(f(x)| $$ shows that $|f(x)| \le 1$ on the intervals $(-3^n \epsilon, 3^n\epsilon)$ for all positive integers $n$, and therefore on all of $\Bbb R$.