Thanks in large part to the commenters for getting me back on the right track. Posting the key points of my solution here for completeness.
$$E[XY]=\int_0^\infty\int_0^yxye^{-y}\,\mathrm dx\,\mathrm dy=3$$
$$E[X]=\int_0^\infty\int_0^yxe^{-y}\,\mathrm dx\,\mathrm dy=1$$
$$E[Y]=\int_0^\infty\int_0^yye^{-y}\,\mathrm dx\,\mathrm dy=2$$
$$E[X^2]=\int_0^\infty\int_0^yx^2e^{-y}\,\mathrm dx\,\mathrm dy=2$$
$$E[Y^2]=\int_0^\infty\int_0^yy^2e^{-y}\,\mathrm dx\,\mathrm dy=6$$
$$\boxed{\mathrm{Cov}[X,Y]=E[XY]-E[X]E[Y]=1}$$
$$\boxed{\mathrm{Corr}[X,Y]=\frac{E[XY]-E[X]E[Y]}{\sqrt{(E[X^2]-E[X]^2)(E[Y^2]-E[Y]^2)}}=\frac1{\sqrt2}}$$
$$\boxed{E[X\mid Y=y]=\int_0^y\frac xy\,\mathrm dx=\frac y2}$$
$$\boxed{E[Y\mid X=x]=\int_x^\infty y^2e^{x-y}\,\mathrm dy=x+1}$$
$$\begin{cases}
E[X]=E[E[X\mid Y]]=E\left[\frac Y2\right]=\frac{E[Y]}2\\[1ex]
E[Y]=E[E[Y\mid X]]=E[X+1]=E[X]+1
\end{cases}\implies\boxed{E[X]=1}$$
$$E[X^2\mid Y=y]=\int_0^y\frac{x^2}y\,\mathrm dx=\frac{y^2}3$$
$$\mathrm{Var}[X\mid Y]=E[X^2\mid Y]-E[X\mid Y]^2=\frac{Y^2}3-\left(\frac Y2\right)^2=\frac{Y^2}{12}$$
$$E[\mathrm{Var}[X\mid Y]]=E\left[\frac{Y^2}{12}\right]=\frac{E[Y^2]}{12}=\frac12$$
$$\mathrm{Var}[E[X\mid Y]]=\mathrm{Var}\left[\frac Y2\right]=\frac{\mathrm{Var}[Y]}4=\frac{E[Y^2]-E[Y]}4=\frac12$$
$$\boxed{\mathrm{Var}[X]=E[\mathrm{Var}[X\mid Y]]+\mathrm{Var}[E[X\mid Y]]=1}$$
Best Answer
When you have $F(x,y)$, you get: $$P(X\le x)=P(X\le x, Y<+\infty)=\lim_{y\to\infty}F(x,y)$$
$$P(Y\le y)=P(X<+\infty,Y\le y)=\lim_{x\to\infty}F(x,y)$$
So you can compute $F_X$ and $F_Y$ passing to the limit.