Compute $E[X \mid Y]$, $E[X]$ and $E[XY^2]$, knowing the distributions of $Y$ and of $X$ conditionally on $Y$ 
Compute $E[X \mid Y = y]$, $E[X]$ and $E[XY^2]$, knowing that $X  \mid Y = y \sim ye^{-yx}$ and $Y \sim e(1)$ and $x,y >0$.

To compute I do as follows: $E[X \mid Y = y] = \int^{\infty}_0 x\cdot ye^{-yx} \ dx = y (\frac {-e^{-xy}(xy+1)} {y^2}]^{\infty}_0 = \frac 1 y$
Then to compute $EX$ i use $E[g(X,Y)] = E[G(Y)]$ where $G(y) := E[g(X,y) \mid Y=y]$ and get $EX = \int^{\infty}_0 \frac 1y e^{-y} \ dy$, but I cannot evaluate this integral ? Also the last expectation is causing me trouble.

One knows that
$$E(X\mid Y)=\frac1Y
$$
hence
$$
E(X)=E(E(X\mid Y))=E\left(\frac1Y\right)=+\infty
$$
and
$$
E(XY^2)=E(E(X\mid Y)Y^2)=E\left(\frac1YY^2\right)=E(Y)=1.$$
Edit: By definition, that the distribution of $Y$ is standard exponential implies (actually, is equivalent to the fact) that, for every nonnegative measurable function $u$,
$$
E(u(Y))=\int_0^\infty u(y)\,\mathrm e^{-y}\,\mathrm dy.
$$
Likewise, by definition, at least for every nonnegative random variable $X$ and nonnegative measurable function $u$,
$$E(Xu(Y))=E(E(X\mid Y)u(Y)).
$$
