jmc

First, we note that $y$ cannot be an integer, since this would imply that $\lceil y\rceil \cdot \lfloor y\rfloor =y^2$, and $42$ is not a perfect square.

Since $y$ is not an integer, we have $\lceil y\rceil =\lfloor y\rfloor +1$. Define $\lceil y\rceil$ as $x$ and $\lfloor y\rfloor$ as $x-1$. If we plug these expressions into the given equation, we get $$\begin{array}{rl}x(x-1)& =42\\ \Rightarrow x^2-x& =42\\ \Rightarrow x^2-x-42& =0\\ \Rightarrow (x-7)(x+6)& =0\end{array}$$This yields $x=7$ and $x=-6$ as the only possible values of $x$. However since the problem states that $y>0$ and $x=\lceil y\rceil$, $x$ must be a positive number and we can eliminate $x=-6$ as a possibility. If $x=\lceil y\rceil =7$, and $x-1=\lfloor y\rfloor =6$, $y$ must be between the integers 6 and 7. Therefore, our final answer is $6<y<7$, which is written in interval notation as $\boxed{(6,7)}$.