If L = {xy | |x| = |y|, x=y} is not Context Free, then what about L = {xy | |x| = |y|, x!=y}?
There is a big difference between $\{ xy ∣ |x|  =  |y|, x = y \}$ and $\{ xy ∣ |x|  =  |y|, x \ne y \}$. In the first one, we need every symbol in $x$ to be the same as the corresponding symbol in $y$. For inequality, it suffices that at least one symbol in x be different from the corresponding symbol in $y$. The two cases are not symmetrical.

Checking that the first symbol is the same is not difficult. That can easily be achieved with a context-free grammar; we just need a word consisting of some symbol followed by a word with that same symbol in the centre:

$$\begin{align}S&amp;\to a A\mid b B\\A&amp;\to a \mid a A a \mid a A b \mid b A a \mid b A b\\B&amp;\to b \mid a B a \mid a B b \mid b B a \mid b B b \\ \end{align}$$

That doesn't help us with checking that all symbols in the first half are the same as the corresponding symbols in the second half. But it does give us a way to check if some symbol in the first half is the same as the corresponding symbol in the second half:

$$\begin{align}S&amp;\to A A\mid B B\\A&amp;\to a \mid a A a \mid a A b \mid b A a \mid b A b\\B&amp;\to b \mid a B a \mid a B b \mid b B a \mid b B b\\  \end{align}$$

And clearly it can easily be modified to check whether the corresponding symbols differ, giving the grammar in @Raphael's answer, linked in your question:

$$\begin{align}S&amp;\to A B\mid B A\\A&amp;\to a \mid a A a \mid a A b \mid b A a \mid b A b\\B&amp;\to b \mid a B a \mid a B b \mid b B a \mid b B b\\  \end{align}$$