S = X + Y .... (1)
T = X/Y .... (2)
$\displaystyle f_{S, T} (s,t) = \frac{\lambda^2 s}{(1 + t)^2} \, e^{-\lambda \, s}$.
From (1) and (2):
X = ST/(1 + T)
Y = S/(1 + T)
$\displaystyle |J| = \frac{s}{(1 + t)^2}$.
Therefore $\displaystyle f_{X,Y} (x,y) = f_{S, T} (s,t)/|J| = \lambda^2 e^{-\lambda s} = \lambda^2 e^{-\lambda (x + y)}$.
This distribution is obviously equivalent to $\displaystyle f_{X,Y} (x,y) = \frac{1}{\lambda^2} e^{-(x + y)/\lambda}$ and there's no problem that I can see.
Edit: I see that you get it now. Excellent. Now the trick is to derive the given joint distribution for S and T from the distributions for X and Y lol!