Yes with smooth $a_{ij}$, say $D_k a_{ij} \in L^\infty(\Omega)$ for every $i,j,k$.

Indeed compute

$div(A\nabla y)$ and $div(A^t\nabla y)$

by owing to the Leibniz rule:

the 1st order parts of both expressions belong to $L^2$ since $y\in H^1$

(a vanishing trace does not help)

and the 2nd order parts are the same since $D_i$ and $D_j$ commute.

On the contrary, the answer is negative if the elements of $A$ are not smooth.

For instance, take a bounded $\Omega\subset R^2$ and $A$ as follows:

$a_{ij}=0$ if $(i,j)\not=(1,2)$ and $a_{12}=a\in L^\infty$ to be specified

(every $a\in L^\infty$ makes the problem meaningful since $y\in H^1$).

Assume now that $y$ is smooth

(also in this case, a vanishing trace does not help).

Then, you can use the Leibniz rule also in this case and obtain

$div(A\nabla y)=D_1(a D_2 y)=a D_1D_2 y+D_1 a D_2 y$

and

$div(A^t\nabla y)=D_2(a D_1 y)=a D_1D_2 y+D_2 a D_1 y$.

Moreover, $a D_1D_2 y\in L^2$ since $y$ is smooth and $a$ is bounded.

Now, if $a$ is smooth with respect to $x_1$ and irregular with respect to $x_2$,

the first expression does belong to $L^2$ while the second one does not.

Did I make some mistake?

Thanks a lot, Gianni. Your arguments sound perfect!