I'm searching for a $3D$ vector field $V$ in $(0,1)^3$ whose parallel projections onto the boundary of $[0,1]^3$ are the following $2D$ vector fields:
The parallel projection of $V$ onto the $x-y$ plane will yield the vector field $\{x\log(x),-y \log(y)\}.$
The parallel projection of $V$ onto the $y-z$ plane will yield the vector field $\{y\log(y),-z\log(z)\}.$
The parallel projection of $V$ onto the $x-z$ plane will yield the vector field $\{x\log(x),-z\log(z)\}.$
I have a good idea of the general shape of the 3D vector field but I haven't been able to express it in mathematical form.
COMMENT.-Just to consider the projection over the X-Y plane. Is it like this wath you want?.
Let the angle $\gamma$ be that of the directional cosinus with the Z-axis and {i,j,k} the base we have $$x\sin(\gamma)i+y\sin(\gamma)j=x\log(x)i-y\log(y)j$$ therefore $\sin(\gamma)=\log(x)$ and $\sin(\gamma)=-\log(y)=\log\dfrac 1y$. Then making $y=\dfrac1x$ you have for $$\sin(\gamma)=\log(x) \text { and } xy=1$$ into the unit cube a solution concerning the X-Y plane.