In the book Mathematical Fallacies, Flaws, and Flimflam, the following fake proof is provided:
The question starts with using trigonometric identities to turn $\sin(3x)\to3\sin(x)-4\sin^3(x)$. Then divides the limit into two where the limit of $({\frac{\sin(x)}{x}})^3$ is obviously $1$, and variable substitutes $x=3t$ making $\frac{\sin(3t)}{(3t)^3}$ which can then be manipulated to become the original equation of $\frac{\sin(3x)}{x^3}$, giving a final result of $-9/2$.
I fail to see where the error is in the calculations, everything seems sound, but I know through the use of L'Hôpital's rule and Desmos that the answer should be $\infty$ at $x=0$.