Since $2011$ is a prime number, every multiple of the left-hand side must be equal to either $1$ or $2011$.
If $x \ge 0$, the first multiple equals $3x$, and there are no solutions. Similarly, there are no solutions if $y \le 0$ (the second multiple is $-3y$).
Suppose now that $x < 0$ and $y > 0$. Then
$$|x+|x+|x|| \cdot ||-y|-y|-y|=|x+|x-x|| \cdot ||y-y|-y|=|x| \cdot |-y| = -xy = -2011.$$
Recalling that $2011$ is prime, we get the above answers.