Given the joint density f(x,y)=\frac{e^{-\frac{x}{y}}e^{-y}}{y} where 0<x,y<\infty. What is the best way to find var(X)?

1) Integrate the joint distribution with respect to y. E.g. \int^{\infty}_{0}\frac{e^{-\frac{x}{y}}e^{-y}}{y}.dy in order to find the marginal distribution of X. Then apply E[X^2]-(E[X])^2 But the above integration is hard to solve.

2) Using conditional probability method. First find the conditional distribution of X given Y=y e.g  (f(X|Y)) and solve for E(X) with E[E(X|Y)]=E[X]. I've got this but what about finding E[X^2] ?

3) Use the fact that Var(X)=E[Var(X|Y)] + Var(E(X|Y)) with Var(X|Y)=E(X^2|Y)-[E(X|Y)]^2 Finding E(X^2|Y) is quite tedious which require to solve for \int^{\infty}_{0}x^2\frac{e^{-\frac{x}{y}}}{y}.dx Moreover, I have no idea how to obtain Var(E(X|Y)) though I know that [tex]E(X|Y)=y[/Math] which is \int^{\infty}_{0}x\frac{e^{-\frac{x}{y}}}{y}.dx

Please help