jmc

As long as $y$ is not an integer, we can define $\lceil y\rceil$ as $x$ and $\lfloor y\rfloor$ as $x-1$. If we plug these expressions into the given equation, we get $$\begin{array}{rl}x(x-1)& =110\\ \Rightarrow x^2-x& =110\\ \Rightarrow x^2-x-110& =0\\ \Rightarrow (x-11)(x+10)& =0\end{array}$$This yields $x=11$ and $x=-10$ as two possible values of $x$. However since the problem states that $y<0$ and $x=\lceil y\rceil$, $x$ cannot be a positive integer. This allows us to eliminate $11$, leaving the $-10$ as the only possible value of $x$. Since $x=\lceil y\rceil =-10$, and $x-1=\lfloor y\rfloor =-11$, $y$ must be between the integers $-10$ and $-11$. Therefore, our final answer is $-11<y<-10,$ or $y\in \boxed{(-11,-10)}$ in interval notation.