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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

For $(y,z)=(xu,x(1-u))$, the inverse isand the Jacobian magnitude of $(y,z)\mapsto (x,u)$ is $1/(y+z)$. Thus the uniform split of a random total has joint densityIntegrating out the other piece yieldsThe transformation $U\mapsto 1-U$ preserves the uniform law, so $Z$ has the same density: