= Solution
Set $U=X/(X+Y)$ and $V=X+Y$. Their inverse <change of variables> is $X=UV$, $Y=(1-U)V$, with domain $0<U<1$, $V>0$. The absolute <Jacobian determinant> of this inverse transformation is
$$
\left|\det\begin{pmatrix}v&u\\-v&1-u\end{pmatrix}\right|=v.
$$
Use the <independence of random variables> to multiply the two original <Gamma distribution> densities. The <change of variables formula> then gives the <joint probability density>
$$
f_{U,V}(u,v)=\frac{\theta^{n+m}}{(n-1)!(m-1)!}\,u^{n-1}(1-u)^{m-1}v^{n+m-1}e^{-\theta v},\qquad 0<u<1,\ v>0.
$$
Factor this into two normalized densities:
$$
f_{U,V}(u,v)=\left[\frac{(n+m-1)!}{(n-1)!(m-1)!}u^{n-1}(1-u)^{m-1}\right]\left[\frac{\theta^{n+m}}{(n+m-1)!}v^{n+m-1}e^{-\theta v}\right].
$$
The support is a product domain, and the factors are precisely a <Beta distribution> density and a <Gamma distribution> density. Therefore
$$
\boxed{U\sim B(n,m),\qquad V\sim\Gamma(n+m,\theta),\qquad U\text{ and }V\text{ independent}.}
$$
This <beta-gamma independence> uses the common rate $\theta$; it is a rate parameter, not the scale parameter $1/\theta$.
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