Is this feynman diagram possible? $e^- + e^+ \to \nu_\mu + \bar{\nu}_\mu$ I am starting to learn about Feynman diagrams and I have been told that flavour has to be conserved at vertices in the case that the exchange particle was neutral. Since here the exchange particle is a charged boson, my guess is that this diagram is possible. Is it true?

 A: I don't see why you "guessed" this. It is impossible by inspection.
Go to your SM Lagrangian and observe that the relevant charged current coupling, the coupling of leptons to the Ws is, broadly, of the form 
$$
\propto W^+_\kappa \bar {\nu}_e \gamma^\kappa (1-\gamma_5) e  +\mathrm{h.c.} ~,
$$
which tells you the Feynman diagram vertex must be of the form "e goes in, and W- and $\nu_e$ go out".
The wrong fictitious superposition of neutrino mass states you are writing, $\nu_\mu$ , is the one coupling to μs in such vertices; in fact, that is how it (neutrino flavor) is defined! 
The convenience superposition of neutrino mass eigenstates coupling to electrons is defined to be $\nu_e$.
(Reviews, such as the PDG (10.4) skip the neutrino flavor label, since it is self-explanatory, and might unnecessarily enhance the illusion these states are real particles.)
A: The process $e^-+e^+\to \nu_{\mu}+\bar{\nu}_{\mu}$ is possible, but via a neutral Z-boson that decays into a truly neutral couple $ \nu_{\mu}+\bar{\nu}_{\mu}$.
A: Even though this question is more than a year old, here is another explanation:
We know that at each vertex, the individual lepton numbers must be conserved,
i.e. $L_e$, $L_{\mu}$ and $L_{\tau}$ are constants.
In the diagram you show, the lepton numbers - both for $L_{e}$ and $L_{\mu}$ - are not conserved. Hence, your diagram cannot be valid.
Does this make sense to you?
