Determine the PDF of $Z = XY$ when the joint pdf of $X$ and $Y$ is given The joint probability density function of random variables $ X$ and $ Y$ is given by 
$$p_{XY}(x,y)=
\begin{cases}
& 2(1-x)\,\,\,\,\,\,\text{if}\,\,\,0<x \le 1, 0 \le y \le 1 \\
& \,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{otherwise.}
\end{cases}
$$
Determine the probbility density function of $ Z = XY$.

There are faster methods, but it can be a good idea, at least once or twice, to calculate the cumulative distribution function, and then differentiate to find the density. 
The upside of doing it that way is that one can retain reasonably good control over what's happening. (There are also a number of downsides!)
So we want $F_Z(z)$, the probability that $Z\le z$. Lets do the easy bits first. It is clear that $F_Z(z)=0$ if $Z\le z$. And it is almost as clear that $F_Z(z)=1$ if $z\ge 1$. So from now on we suppose that $0\lt z\lt 1$.
Draw a picture of our square. For fixed $z$ between $0$ and $1$, draw the first quadrant part of the curve with equation $xy=z$.
This curve is a rectangular hyperbola, with the positive $x$ and $y$ axes as asymptotes.  We want the probability that $(X,Y)$ lands in the part of our square which is on the "origin side" of the hyperbola.
So we need to integrate our joint density function over this region. There is some messiness in evaluating this integral: we need to break up the integral at $x=z$. We get   
$$F_Z(z)= \Pr(Z\le  z)=\int_{x=0}^z \left(\int_{y=0}^1 (2-2x)\,dy\right)\,dx +  \int_{x=z}^1 \left(\int_{y=0}^{z/x} (2-2x)\,dy\right)\,dx. $$
Not difficult after that. We get, I think, $F_Z(z)=z^2-2z\ln z$.  
Differentiate for the density. 
