$X$ and $Y$ have a joint distribution with PDF $f(x,y) = e^{-(x + y)} , x > 0 , y > 0$. Given $U = e^{-(X + Y)}$, find the PDF of $U$.
What I did :
Clearly, $X$ and $Y$ are i.i.d $\exp(1)$ , so $X+Y$ ~ $\operatorname{Gamma}(2,1)$. Let $Z$ = $X+Y$.
$$P(U < u) = P(e^{-z} < u) = P(Z > -\ln(u))$$ which after solving I get $\dfrac{1}{u}(1 - \ln(u) ).$
But answer to this question is that $U$'s PDF is $-\ln(u)$. Where am I wrong ? Kindly help.
Without seeing how you did the calculation, it's not possible to further pinpoint where you went wrong, but $$\Pr[Z > -\log u] \ne \frac{1}{u}(1 - \log u).$$
You are correct that $Z = X+Y \sim \operatorname{Gamma}(2,1)$. This gives us the PDF $$f_Z(z) = z e^{-z}, \quad z > 0.$$ The survival function is $$S_Z(z) = \Pr[Z > z] = \int_{t=z}^\infty t e^{-t} \, dt = \left[-(1+t) e^{-t} \right]_{t=z}^\infty = (1+z)e^{-z}.$$ This in turn implies $$F_U(u) = \Pr[U \le u] = \Pr[Z \ge -\log u] = u (1 - \log u), \quad 0 < u \le 1,$$ from which the density is simply $$f_U(u) = \frac{d}{du}\left[u (1 - \log u)\right] = -\log u,$$ as claimed.
That said, this calculation is unnecessary: because $u = g(z) = e^{-z}$ is a monotone transformation, it directly follows that
$$f_U(u) = f_Z(g^{-1}(u))\left|\frac{dg^{-1}}{du}\right| = (-\log u) e^{-(-\log u)} \left|\frac{d}{du}[-\log u]\right| = - \log u.$$