For solving, you can notice that if $x<0$, both the terms in LHS will be negative, and therefore cann't add up to $0$. If $x>0$ then both terms in LHS will be non-negative. Hence, they'll sum up to $0$ only when both the terms are $0$ individually. I.e. $$\lfloor x\rfloor =\lfloor 3x\rfloor =0 \implies x \in [0,1) \cap [0,1/3)$$
$$\implies x \in [0,1/3)$$
What you've done is correct, but you don't get the complete solution set for the given equation, because the inequality used by you isn't "strong" enough.
What you get from this is a 'superset' of the complete solution.