The correct answer is $\Theta(n \log n)$.
To see why, we rewrite the recurrence relation to avoid $\Theta$ notation as follows:
$$T(x, y)=c x+T(x, y / 2) .$$ We may then begin to replace $T(x, y / 2)$ with the recursive formula containing it:
$$T(x, y)=\underbrace{c x+c x+c x+\ldots+c x}_{\Theta(\log y) \text { times }} .$$ As a result, $T(x, y)$ is $\Theta(x \log y)$.
When we substitute $n$ for $x$ and $y$, we get that $T(n, n)$ is $\boxed{\Theta(n \log n)}$
