As usual with maxima/minima, the event $A_{u,v}=[v\leqslant V,U\leqslant u]$ for $v\leqslant u$ is well suited to the computation of the distribution of $(U,V)$ since $A_{u,v}=[v\leqslant X,Y,Z\leqslant u]$, hence $ \mathbb P(A_{u,v})=\int\!\!\!\iint f(x,y,z)\,\mathbf 1_{v\leqslant x,y,z\leqslant u}\,\mathrm dx\mathrm dy\mathrm dz. $ Due to the symmetries of $f$, this is $ \mathbb P(A_{u,v})=6\int_v^u\int_v^x\int_v^yf(x,y,z)\,\mathrm dz\mathrm dy\mathrm dx, $ hence $ \partial_u\mathbb P(A_{u,v})=6\int_v^u\int_v^yf(u,y,z)\,\mathrm dz\mathrm dy=6\int_v^u\int_z^uf(u,y,z)\,\mathrm dy\mathrm dz, $ and $ \partial_v\partial_u\mathbb P(A_{u,v})=-6\int_v^uf(u,y,v)\,\mathrm dy. $ This yields the density $f_{U,V}$ of $(U,V)$ as, for example, $ f_{U,V}(u,v)=6\int_v^uf(u,v,w)\,\mathrm dw\cdot\mathbf 1_{u\leqslant v}, $ which has the following infinitesimal interpretation: to get $U\approx u$ and $V\approx v$, order the sample ($6$ choices), put the lowest value of the sample at $u$, the highest value at $v$ and the middle value at $w$ somewhere in $(u,v)$.