If $F^{\alpha \beta} = -F^{\beta \alpha}$ (i.e. anti-symmetric) then prove the following:
$F^{\alpha}_{\mu,\nu} F^{\nu}_{\alpha} = - F_{\mu\alpha,\beta}F^{\alpha\beta}$
My proof is as follows (where $g_{\sigma\rho}$ is the metric), \begin{align*} F^{\alpha}_{\mu,\nu} F^{\nu}_{\alpha} &= \frac{\partial g^{\alpha\beta} F_{\mu\beta}}{\partial x^{\nu}}F^{\nu}_{\alpha} \\ &= \frac{\partial F_{\mu\beta}}{\partial x^{\nu}}g^{\alpha\beta} F^{\nu}_{\alpha} \\ &= \frac{\partial F_{\mu \beta}}{\partial x^{\nu}}F^{\beta \nu}\\ &= -F_{\mu \beta,\nu}F^{\nu\beta} \end{align*}
My main concern is whether or not I can just move the metric out of the derivative like that, as well as my indices having different labels to the identity (i.e. I didn't change $\nu$ to $\beta$ in the partial derivative)?
I worked through the problem a few more times and this is my new answer,
$\begin{align} F_{\mu,\nu}^{\alpha}F_{\alpha}^{\nu} &= \frac{\partial g^{\alpha \sigma} F_{\sigma\mu}}{\partial x^{\nu}}g_{\sigma\alpha}F^{\nu\sigma} \\ &= g^{\alpha \sigma}_{,\nu}F_{\sigma\mu}g_{\sigma\alpha}F^{\nu\sigma} + g^{\alpha \sigma}F_{\sigma\mu,\nu}F^{\nu\sigma}g_{\sigma\alpha} \end{align}$
but if the metric is constant (e.g. the Minkowski metric) then its derivative is zero thus,
$=g_{\sigma\alpha}g^{\alpha \sigma}F_{\sigma\mu,\nu}F^{\nu\sigma}$
Since the metrics become the identity matrix, and since the tensor $F$ is anti-symmetric,
$=-F_{\sigma\mu,\nu}F^{\sigma\nu}$