d’Alembert Lagrangian for second rank tensors 
Consider the Lorentz scalar Lagrange density
  $$\mathcal{L}=\eta^{\mu\nu}\partial_\mu T^{\alpha\beta}\partial_\nu T_{\alpha\beta}$$
  for a second rank tensor whose contravariant and covariant components are $T^{\alpha\beta}$, $T_{\alpha\beta}$ respectively. By varying with respect to each component $T^{\alpha\beta}$ separately, show that the set of generalized Euler-Lagrange equations for $\mathcal{L}$ give the set of equations of motion
  $$\square^2T^{\alpha\beta}=0, \quad \text{for } \alpha,\beta=0,1,2,3.$$

In the previous statement, we have that $\eta^{\mu\nu}$ is the metric tensor defined as 
$$\eta^{\mu\nu}\left\{
\begin{array}{ccl}
1 &\text{if}& \nu=\mu=0, \\
-1 &\text{if}& \nu=\mu=1,2,3, \\
1 &\text{if}& \nu\neq\mu, \\
\end{array}
\right. $$
and $\partial_\mu$ is the $4-$gradient given by 
$$\partial_\mu=\frac{\partial }{\partial x^\mu}.$$
Firstly, we can express the Lagrange density as 
$$\mathcal{L}=\eta^{\mu\nu}\partial_\mu T^{\alpha\beta}\partial_\nu T_{\alpha\beta}=\eta^{\mu\nu}\eta^{\alpha\rho}\eta^{\beta\sigma}\partial_\mu T_{\rho\sigma}\partial_\nu T_{\alpha\beta}.$$
We want to find an expression of the Euler-Lagrange equations
$$\frac{\partial \mathcal{L}}{\partial T_{\alpha\beta}}-\partial_\mu \left(\frac{\partial \mathcal{L}}{\partial (\partial_\mu T_{\alpha\beta})}\right)=0, \quad \text{ for } \alpha,\beta=0,1,2,3,$$
in terms of our Lagrange density. Observe that, since $\mathcal{L}$ does not depend explicitly on $T_{\alpha\beta}$ (the dependance is on their derivatives), the first term is zero. On the other hand, 
$$\frac{\partial \mathcal{L}}{\partial (\partial_\mu T_{\alpha\beta})}=\frac{\partial }{\partial (\partial_\mu T_{\alpha\beta})}\left(\partial_\mu T_{\rho\sigma}\partial_\nu T_{\alpha\beta}\right)=\frac{\partial(\partial_\mu T_{\rho\sigma})}{\partial(\partial_\mu T_{\alpha\beta})}\partial_\nu T_{\alpha\beta}+\frac{\partial(\partial_\nu T_{\alpha\beta})}{\partial(\partial_\mu T_{\alpha\beta})}\partial_\mu T_{\rho\sigma}.$$
However, I do not know how to proceed from here because this index notation still confuses me. The result should be $2\partial^\mu T^{\alpha\beta}$ and, if I arrive to that point, I know how to conclude the problem. 
Can anyone explain to me how these derivatives give the right answer? I insist that this is the first time I encounter the sum Einstein convention. Thank you in advance. 
 A: Having said that this is not the way I would carry out a variation, you are almost there. Take $\partial_{\alpha}T_{\mu \nu}=T_{\mu \nu,\alpha}$, then 
$$
\frac{\partial \mathcal{L}}{\partial T_{\mu \nu,\alpha}}=\eta^{\rho \sigma}\eta^{\lambda \beta} \eta^{\tau \gamma} \frac{\partial }{\partial T_{\mu \nu,\alpha}} (T_{\lambda \tau,\rho}T_{\beta \gamma,\sigma})=\eta^{\rho \sigma}\eta^{\lambda \beta} \eta^{\tau \gamma} (\delta^{\mu}_{\lambda}\delta^{\nu}_{\tau}\delta^{\alpha}_{\rho}T_{\beta \gamma,\sigma}+\delta^{\mu}_{\beta}\delta^{\nu}_{\gamma}\delta^{\alpha}_{\sigma}T_{\lambda \tau,\rho})
$$
if you use that $\eta$ is symmetric in its indeces, you can change the index name of one term in the parenthesis so as to add it to the other and get
$$
\frac{\partial \mathcal{L}}{\partial T_{\mu \nu,\alpha}}=2\eta^{\mu \beta}\eta^{\nu \gamma}\eta^{\alpha \sigma}T_{\beta \gamma,\sigma}
$$ 
then 
$$
0=\partial_{\alpha}\frac{\partial \mathcal{L}}{\partial T_{\mu \nu,\alpha}}=2\eta^{\mu \beta}\eta^{\nu \gamma}\eta^{\alpha \sigma}\partial_{\alpha}\partial_{\sigma}T_{\beta \gamma}=2\eta^{\mu \beta}\eta^{\nu \gamma} \square (T_{\beta \gamma}).
$$
Let me remark a few things: first this derivation is completely general so that you can take any metric $g$ in place of $\eta$, second when you use indeces try not to repeat those in formulas when making substitution like you did in the calculation of $\frac{\partial \mathcal{L}}{\partial T_{\mu \nu,\alpha}}$ since when you use Einstein convention (i.e. repeated indeces are summed over) then
$$ 
\frac{\partial T_{\mu \nu,\alpha}}{\partial T_{\mu \nu,\alpha}} = \sum_{\mu,\nu,\alpha =0}^3\delta^{\mu}_{\mu}\delta^{\nu}_{\nu}\delta^{\alpha}_{\alpha}=4\cdot 4\cdot 4=64.
$$
In general a good way to check the notational correctedness of your formula is to ensure that to each free index (i.e. not repeated index) in the left hand side there is the same free index in the right hand side. 
