If I have an interaction term in my Lagrangian that looks like:
$\mathcal{L}_{int} = (\partial_\mu B_\nu)(A^\mu B^\nu - A^\nu B^\mu)$ where B is a massive spin-1 field.
Am I correct in thinking that the vertex rule associated with this term is 0? Since the vertex rule would be proportional to $k_B^\mu (g^{\nu \rho} - g^{\rho \nu})$ = 0
The vertex generated by your lagrangian is nonzero.
Suppose you have, by definition, that $$ \tag 1 \frac{\delta A^{\nu}}{\delta A_{\mu}} \equiv g^{\mu \nu} $$ In Fourier space the 3-vertex generated by your largangian is obtained by "amputation" of fields by using $(1)$, $$ \Gamma^{\alpha \beta \delta} \sim \frac{\delta^{2}}{\delta B_{\alpha}\delta B_{\beta}}\frac{\delta}{\delta A_{\gamma}}\left(k_{\mu}B_{\nu}(A^{\nu}B^{\mu} - A^{\mu}B^{\nu})\right) = $$ $$ =\frac{\delta^{2}}{\delta B_{\alpha}\delta B_{\beta}}k_{\mu}B_{\nu}\left(g^{\gamma \nu}B^{\mu} - g^{\gamma \mu}B^{\nu} \right) = $$ $$ =2k_{\mu}\left( g^{\beta \gamma}g^{\alpha \mu} - g^{\gamma \mu}g^{\alpha \beta} - g^{\alpha \gamma}g^{\beta \mu} + g^{\gamma \mu}g^{\alpha \beta}\right) = $$ $$ = 2(k^{\alpha}g^{\beta \gamma} - k^{\beta}g^{\alpha \gamma}) $$
