R random vector generator 
Create an R function generating ordered pairs x,y sampled from the two
  dimensional distribution whose pdf is of the form $f(x,y)=cxy$, where
  $0<x,y<1$, and $c$ is a constant to be determined.

Integrating the pdf, I can determine that:
$$\int\limits_0^1 \int\limits_0^1 c \, xy \, dx \, dy = 1 \Rightarrow c=4$$
Edit:
@Flounderer already suggested me how to create a random vector generator, but need to create a different one for the same density, what should I do?
 A: Three general approaches are described in the answer to this question but in your case it's easier. 
You already found the joint density $f(x,y) = 4xy$. It factors into a function of $x$ and a function of $y$, so $X$ and $Y$ are independent. Therefore, you just need to generate $X$ and $Y$ separately. You can find their marginal distributions by integrating out to get $f_X(x) = 2x$ and $f_Y(y) = 2y$. To generate a random number from the distribution with pdf $f_X(x) = 2x$, you can generate a uniform random number and apply the inverse cdf. In this case, the cdf is $F_X(x) = \int_0^x f_X(t) dt = x^2$, so you can generate a random number from this distribution by doing $F_X^{-1}(U) = \sqrt{U}$ where $U \sim U(0,1)$. Similarly, you can generate a value of $Y$. Overall, $(X,Y) = (\sqrt{U}, \sqrt{V})$ where $U, V \sim U(0,1)$ are independent. 
In R, a random vector from your distribution can be generated as 
sqrt(runif(2))

To generate a matrix whose rows are $n$ random vectors sampled from your distribution you can do
sqrt(matrix(runif(n*2), ncol=2))

You can check graphically that this works.
A: You've determined $c$, so you know the full pdf of the joint distribution of $(X, Y)$.  Now:


*

*Determine the pdf of the conditional distribution of $Y|X$, this amounts to finding the normalization factor you need for each value of X.

*Determine the pdf of the distribution of $X$.  This involves integrating out the effect of $Y$.

*First sample X, then sample Y conditioned on X.


Note that you cannot assume that $X$ and $Y$ are $U[0,1]$, though you will discover what they are by going through the procedure above.
