From what I gather from your latest comments, the PDF is given to be

and in particular, <em>f(x, y)</em> = <em>cxy</em> over the unit square [0, 1]², meaning for 0 ≤ <em>x</em> ≤ 1 and 0 ≤ <em>y</em> ≤ 1. (As opposed to the unbounded domain, <em>x</em> ≤ 0 *and* <em>y</em> ≤ 1.)
(a) Find <em>c</em> such that <em>f</em> is a proper density function. This would require

(b) Get the marginal density of <em>X</em> by integrating the joint density with respect to <em>y</em> :

(c) Get the marginal density of <em>Y</em> by integrating with respect to <em>x</em> instead:

(d) The conditional distribution of <em>X</em> given <em>Y</em> can obtained by dividing the joint density by the marginal density of <em>Y</em> (which follows directly from the definition of conditional probability):

(e) From the definition of expectation:
![E[X]=\displaystyle\int_0^1\int_0^1 x\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\boxed{\frac23}](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cint_0%5E1%5Cint_0%5E1%20x%5C%2Cf_%7BX%2CY%7D%28x%2Cy%29%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%3D4%5Cleft%28%5Cint_0%5E1x%5E2%5C%2C%5Cmathrm%20dx%5Cright%29%5Cleft%28%5Cint_0%5E1y%5C%2C%5Cmathrm%20dy%5Cright%29%3D%5Cboxed%7B%5Cfrac23%7D)
![E[Y]=\displaystyle\int_0^1\int_0^1 y\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac23}](https://tex.z-dn.net/?f=E%5BY%5D%3D%5Cdisplaystyle%5Cint_0%5E1%5Cint_0%5E1%20y%5C%2Cf_%7BX%2CY%7D%28x%2Cy%29%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%3D4%5Cleft%28%5Cint_0%5E1x%5C%2C%5Cmathrm%20dx%5Cright%29%5Cleft%28%5Cint_0%5E1y%5E2%5C%2C%5Cmathrm%20dy%5Cright%29%3D%5Cboxed%7B%5Cfrac23%7D)
![E[XY]=\displaystyle\int_0^1\int_0^1xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac49}](https://tex.z-dn.net/?f=E%5BXY%5D%3D%5Cdisplaystyle%5Cint_0%5E1%5Cint_0%5E1xy%5C%2Cf_%7BX%2CY%7D%28x%2Cy%29%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%3D4%5Cleft%28%5Cint_0%5E1x%5E2%5C%2C%5Cmathrm%20dx%5Cright%29%5Cleft%28%5Cint_0%5E1y%5E2%5C%2C%5Cmathrm%20dy%5Cright%29%3D%5Cboxed%7B%5Cfrac49%7D)
(f) Note that the density of <em>X</em> | <em>Y</em> in part (d) identical to the marginal density of <em>X</em> found in (b), so yes, <em>X</em> and <em>Y</em> are indeed independent.
The result in (e) agrees with this conclusion, since E[<em>XY</em>] = E[<em>X</em>] E[<em>Y</em>] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)