Divergenceless proving We can define a topological current,
\begin{equation}
J_{top}^u = \frac{1}{2v} \epsilon^{\mu \nu} \partial_\nu \phi
\end{equation}
How to prove it divergenceless? where the the condition is $$\partial_uJ=0.$$
 A: In answering this question, I am going to make a few assumptions which clarify the OP's statements and (hopefully) refine some of the definitions, both implicit and explicit, inherent in his post.  I am going to go ahead and do this without resorting to the process of commenting and waiting for responses before answering; with luck, I'll hit the mark.
First of all, I'm taking the first equation, which defines $J$ (and please note I'm dropping the subscript "$top$" here), to read
$J^\mu = \frac{1}{2v}\epsilon^{\mu \nu} \partial_\nu \phi$,
and the second to be
$\partial_\mu J^\mu = 0$.
These assumptions merely correct what appear to be typographical errors, and make the indexing scheme consistent.  Finally, I take $\epsilon^{\mu \nu}$ to be an antisymmetric symbol on two indices, meaning
1.)  $\epsilon^{\mu \nu} = 0$ for $\mu = \nu$,
2.)  $\epsilon^{\mu \nu} = (-1)^{\mu + \nu + 1}$ for $\nu > \mu$,
3.)  $\epsilon^{\mu \nu} = -\epsilon^{\nu \mu}$ for $\nu < \mu$;
where it is assumed the indices run from $1$ to $N$, where $N$ is the dimension of the space under consideration.  With these conventions, and assuming $v$ is constant and $\phi$ is sufficiently smooth, so that $\partial_\mu \partial_\nu \phi =  \partial_\nu \partial_\mu \phi$, we have
$\partial_\alpha J^\mu = \frac{1}{2v}\epsilon^{\mu \nu} \partial_\alpha \partial_\nu \phi$;
then contracting on the indices $\alpha$, $\mu$, i.e., taking $\alpha = \mu$ and summing yields
$\partial_\mu J^\mu = \frac{1}{2v}\epsilon^{\mu \nu} \partial_\mu \partial_\nu \phi = -\frac{1}{2v}\epsilon^{\nu \mu} \partial_\mu \partial_\nu \phi = -\frac{1}{2v}\epsilon^{\nu \mu} \partial_\nu \partial_\mu \phi = -\partial_\nu J^\nu = -\partial_\mu J^\mu$,
whence 
$\partial_\mu J^\mu = 0$,
since we are not operating over a field of characteristic 2!  
Hope this helps.  Cheers.
