I guess there is a typo and the question is to show

$$\partial_\mu \frac{\partial (\partial_\alpha A_\alpha)^2}{\partial (\partial_\mu A_\nu)} = \partial_\mu \left[2 (\partial_\alpha A_\alpha)\frac{\partial  (\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)} g_{\beta\gamma}\right]\ .$$

As mentioned in the OP's [link][1] Schwartz doesn't keep track of the index placement on tensor objects that obscures the structure a little. 

However, ignoring the first derivative $\partial_\mu$ the starting point is to distinguish between covariant and contravariant indices such that 
$$ (\partial_\alpha A_\alpha)^2 \to (\partial_\alpha A^\alpha)^2=(\partial_\alpha A^\alpha)(\partial_\beta A^\beta)\ .$$

Applying the product rule we see
$$\frac{\partial (\partial_\alpha A^\alpha)(\partial_\beta A^\beta)}{\partial (\partial_\mu A_\nu)}= \frac{\partial (\partial_\alpha A^\alpha)}{\partial (\partial_\mu A_\nu)}(\partial_\beta A^\beta)+(\partial_\alpha A^\alpha)\frac{\partial(\partial_\beta A^\beta)}{\partial (\partial_\mu A_\nu)}\ .$$

Since the indices must be at the right position in order to perform a derivative we can use the metric to write $\partial_\alpha A^\alpha= g^{\alpha\gamma}\partial_\alpha A_\gamma$ similar for the second expression. 

Hence, the Euler-Lagrange variation involves a metric
$$\frac{\partial (\partial_\alpha A^\alpha)}{\partial (\partial_\mu A_\nu)}=\frac{\partial (\partial_\alpha A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\alpha\gamma}\ .$$

Collecting all pieces and replacing the labeling of dummy indices $\alpha \leftrightarrow \beta$ this leads eventually to the result in the brackets
$$\frac{\partial (\partial_\alpha A^\alpha)(\partial_\beta A^\beta)}{\partial (\partial_\mu A_\nu)} = (\partial_\beta A^\beta)\frac{\partial (\partial_\alpha A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\alpha\gamma}+(\partial_\alpha A^\alpha)\frac{\partial(\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta \gamma}\\
=(\partial_\alpha A^\alpha)\frac{\partial (\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta\gamma}+(\partial_\alpha A^\alpha)\frac{\partial(\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta \gamma}\\
=2 (\partial_\alpha A^\alpha)\frac{\partial (\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta\gamma}\ .$$


  [1]: https://physics.stackexchange.com/questions/355018/how-come-frac-partial-partial-betaa-gamma-partial-partial-mua