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From what I gather from your latest comments, the PDF is given to be
[tex]f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0\le x,y \le1\\0&\text{otherwise}\end{cases}[/tex]
and in particular, f(x, y) = cxy over the unit square [0, 1]², meaning for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (As opposed to the unbounded domain, x ≤ 0 *and* y ≤ 1.)
(a) Find c such that f is a proper density function. This would require
[tex]\displaystyle\int_0^1\int_0^1 cxy\,\mathrm dx\,\mathrm dy=c\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\frac c{2^2}=1\implies \boxed{c=4}[/tex]
(b) Get the marginal density of X by integrating the joint density with respect to y :
[tex]f_X(x)=\displaystyle\int_0^1 4xy\,\mathrm dy=(2xy^2)\bigg|_{y=0}^{y=1}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}[/tex]
(c) Get the marginal density of Y by integrating with respect to x instead:
[tex]f_Y(y)=\displaystyle\int_0^14xy\,\mathrm dx=\begin{cases}2y&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}[/tex]
(d) The conditional distribution of X given Y can obtained by dividing the joint density by the marginal density of Y (which follows directly from the definition of conditional probability):
[tex]f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}[/tex]
(e) From the definition of expectation:
[tex]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}[/tex]
[tex]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}[/tex]
[tex]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}[/tex]
(f) Note that the density of X | Y in part (d) identical to the marginal density of X found in (b), so yes, X and Y are indeed independent.
The result in (e) agrees with this conclusion, since E[XY] = E[X] E[Y] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)