So give -ln(X) $\sim$ exp(1) which is well established where X is $U(0,1)$, then similarly -ln(Y) $\sim exp(1)$ where Y is uniform.
Then sums of exponentials are gamma(2,1) indicates that -ln(X)-ln(Y) $\sim Gamma(2,1)$
$f_{W}(y)$= $ye^{-y}, y\geq 0$
then the joint distribution $f_{Z,W} = ye^{-y}, y \geq 0$ since Z $\sim U(0,1)$
now the joint distribution is established.
if you solve for the transformation U=ZW, and V=$\frac{W}{Z}$, set up your jacobian and then find the marginal of U by integrating over all V and you should get your answer which is an exponential term.