Let $g(x)=f(x+1)$, then $g(3x)\le g(x)\le g(2x)$, therefore if g(x) continiosly in 0, then $g(x)\equiv const$.

can you explain your solution,please?

we can easily show by induction that : $ g(\frac{x}{2^{n}})\leq\ g(x)\leq\ g(\frac{x}{3^{n}}) $ , $ \forall\ (x,n)\in\mathbb{R}\times\mathbb{N} $ now for a fixed $x$ take the limit when $n\mapsto\infty$ using the fact that $f$ is continuous we get : $g(x)=g(0)$ , $\forall\ x\in\mathbb{R} $ so: $g$ and then $f$ are constant!

Another problem of this type, which includes this particular one as a special case (for the function $g$) has already been posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=379114&hilit=3x