I was able to calculate the density of the $XY$. My attempt went like this: if we consider the intervals in which $X$ and $Y$ are equal to $1$ $(0 < x,y < 1)$ then the density of $XY$ is $1$ when $0 < xy < 1$ and $0$ for all other values of $xy$. Finding the cdf from this was much trickier however, I realize that you have to integrate the density, but I'm not sure with respect to what.
For $X/Y$, it is much trickier to calculate the density because there are several cases we have to consider. $Y$ cannot equal $0$ otherwise $X/Y$ is undefined (would this be written as $f(X/Y) = 0$ if $Y = 0$)? This second part is much more confusing, so I need help finding both the cdf and pdf of this.
No need to worry about $Y=0$, or any other particular value of $Y$, as particular values have probability zero.
The cdf of $X/Y$ is given by $$F(z)=P(X/Y\le z) = P(X\le zY)=\iint_{x\le zy} dx\,dy$$ and then the pdf is the derivative $F'(z)$.