A question of Joint PDF 
I have not idea about part a. I know I need to prove the integration of f(x,y)=1, but how should I deal with the range of x and y.
 A: We deal with part a), which is causing you some difficulty. It is basically a calculus problem. 
Our density function is $0$ except on the part of the $x$-$y$ plane which is below the line $y=x$. Call this region $D$. We want to show that 
$$\iint_D \frac{2e^{-2x}}{x}\,dx\,dy=1.$$
One standard way to evaluate this kind of double integral is to express it as an iterated integral. We have in principle a choice as to whether to integrate first with respect to $x$ or with respect to $y$. With respect to $x$ looks difficult. So let us integrate first with respect to $y$. We get
$$\int_{x=0}^\infty \left(\int_0^x \frac{2e^{-2x}}{x}\,dy\right)\,dx.$$
The inner integral is simply $x\frac{2e^{-2x}}{x}$, which is $2e^{-2x}$. The outer integral is therefore 
$$\int_{0}^\infty 2e^{-2x}\,dx.$$
This integral is easy to evaluate. It is $1$.
On the way, we have found the (marginal) density of $X$, for that is obtained by "integrating out" $y$. The marginal density is $2e^{-2x}$ (when $x\gt 0$). For $x\le 0$, the density function of $X$ is $0$.
