While "independent" specifies a precise distribution of the pair $(X,Y)$, the word "dependent" does not - that is, merely knowing that $X$ and $Y$ are both individually uniform is not actually that helpful.
You can still get the answer of $1/2$ in a fairly wide range of cases - if $(X,Y)$ and $(-X,-Y)$ follow the same distribution (i.e. there's a negation symmetry) and the probability that $X=3Y$ is $0$, then it follows that $X<3Y$ and $X>3Y$ are equally likely, so happen with probability $1/2$. However, in general, this cannot be said.
You can be sure of some things: if $Y\geq 1/3$, then $3Y\geq X$ because $X\leq 1$. Similarly, if $Y<-1/3$, then $3Y<X$. Thus, the probability that $X>3Y$ is at least $1/3$ and no more than $2/3$ in any case. Any probability between those is fair game, however. For instance, you could set up a distribution as follows:
Choose $Y$ uniformly at random from $[-1,1]$. If $|Y|<1/3$, then set $X=3Y$. Otherwise, choose $X$ uniformly at random from $[-1,1]$.
You can verify that $X$ is uniformly distributed on $[-1,1]$ but $P(X>3Y)=1/3$. At the other extreme, you can do the following:
Choose $Y$ uniformly at random from $[-1,1]$. If $|Y|<1/3$, then set $X=Y+2/3$. Otherwise, choose $X$ uniformly at random from $[-1,1/3]$.
It's a little harder to see that $X$ is still uniform - basically, you split into cases - given that $|Y|$ was less than $1/3$, then $X$ is uniform random on $[1/3,1]$ and otherwise it is uniform on $[-1,1/3]$. The probabilities of these cases work out to give a uniform distribution across the whole interval. Once you verify that, you find that $P(X>3Y)=2/3$ under this distribution - and modifying the translation of $X$ from $Y$ between $+2/3$ and $-2/3$ actually gives every feasible probability.
Basically, there are a lot of ways to make distributions that are not independent, so only the "obvious" constraints remain when you say merely "not independent"