= Solution
The map $(x,u)\mapsto(x,y)=(x,xu)$ has inverse $u=y/x$ and Jacobian magnitude $1/x$. Hence the density of $(X,Y)$ is
$$
\boxed{f_{X,Y}(x,y)=\frac{g(x)}x
\mathbf 1_{\{x>0,\,0\leq y\leq x\}}}.
$$
For $(y,z)=(xu,x(1-u))$, the inverse is
$$
x=y+z,
\qquad
u=\frac y{y+z},
$$
and the Jacobian magnitude of $(y,z)\mapsto(x,u)$ is $1/(y+z)$. Thus the <uniform split of a random total> has joint density
$$
\boxed{f_{Y,Z}(y,z)=\frac{g(y+z)}{y+z}
\mathbf 1_{\{y,z\geq0\}}}.
$$
Integrating out the other piece yields
$$
\boxed{f_Y(y)=\int_0^\infty\frac{g(y+z)}{y+z}\,dz=h(y)},
\qquad y\geq0.
$$
The transformation $U\mapsto1-U$ preserves the uniform law, so $Z$ has the same density:
$$
\boxed{f_Z(z)=h(z)},
\qquad z\geq0.
$$
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