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 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 $$\frac{\partial (\partial_\alpha A^\alpha)(\partial_\beta A^\beta)}{\partial (\partial_\mu A_\nu)} = (\partial_\beta A^\beta)\frac{\partial (\partial_\alpha A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\alpha\gamma}+(\partial_\alpha A^\alpha)\frac{\partial(\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta \gamma}\\ =(\partial_\alpha A^\alpha)\frac{\partial (\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta\gamma}+(\partial_\alpha A^\alpha)\frac{\partial(\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta \gamma}\\ =2 (\partial_\alpha A^\alpha)\frac{\partial (\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta\gamma}\ .$$

As mentioned in the OP's link Schwartz (in his QFT book) doesn't keep track of the index placement on tensor objects that might obscure the structure a little.

However, ignoring the first derivative $\partial_\mu$ (because it is not relevant for the answer) 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 $$\frac{\partial (\partial_\alpha A^\alpha)(\partial_\beta A^\beta)}{\partial (\partial_\mu A_\nu)} = (\partial_\beta A^\beta)\frac{\partial (\partial_\alpha A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\alpha\gamma}+(\partial_\alpha A^\alpha)\frac{\partial(\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta \gamma}\\ =(\partial_\alpha A^\alpha)\frac{\partial (\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta\gamma}+(\partial_\alpha A^\alpha)\frac{\partial(\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta \gamma}\\ =2 (\partial_\alpha A^\alpha)\frac{\partial (\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta\gamma}\ .$$

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 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 $$\frac{\partial (\partial_\alpha A^\alpha)(\partial_\beta A^\beta)}{\partial (\partial_\mu A_\nu)} = (\partial_\beta A^\beta)\frac{\partial (\partial_\alpha A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\alpha\gamma}+(\partial_\alpha A^\alpha)\frac{\partial(\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta \gamma}\\ =(\partial_\alpha A^\alpha)\frac{\partial (\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta\gamma}+(\partial_\alpha A^\alpha)\frac{\partial(\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta \gamma}\\ =2 (\partial_\alpha A^\alpha)\frac{\partial (\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta\gamma}\ .$$

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 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 $$\frac{\partial (\partial_\alpha A^\alpha)(\partial_\beta A^\beta)}{\partial (\partial_\mu A_\nu)} = (\partial_\beta A^\beta)\frac{\partial (\partial_\alpha A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\alpha\gamma}+(\partial_\alpha A^\alpha)\frac{\partial(\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta \gamma}\\ =(\partial_\alpha A^\alpha)\frac{\partial (\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta\gamma}+(\partial_\alpha A^\alpha)\frac{\partial(\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta \gamma}\\ =2 (\partial_\alpha A^\alpha)\frac{\partial (\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta\gamma}\ .$$