The divergence of a vector field \boldsymbol{u}=\pmatrix{u_1,u_2,u_3} is given by
\boldsymbol{\nabla\cdot u}=\frac{\partial u_1}{\partial x}+\frac{\partial u_2}{\partial y}+\frac{\partial u_3}{\partial z}.
a)
The variables x, y, and z appear in a cyclical manner in each of the three components of \boldsymbol{u}. Once you have calculated \frac{\partial u_1}{\partial x}, you can use cyclic permutations to determine the other two derivatives. Hence
\begin{align}\frac{\partial u_1}{\partial x}&=\frac{\partial}{\partial x}\left(\simplify{({a1}*x+{b1}*y+{c1}*z)*({b1}*y-{c1}*z)}\right)\\&=\simplify{{a1}*({b1}*y-{c1}*z)},\end{align}
and so, cyclically permuting the variables,
\frac{\partial u_2}{\partial y}=\simplify{{a1}*({b1}*z-{c1}*x)}
and
\frac{\partial u_3}{\partial z}=\simplify{{a1}*({b1}*x-{c1}*y)}.
Finally, adding the components together gives the divergence
\boldsymbol{\nabla\cdot u}=\simplify[std]{{a1}*({b1}*y-{c1}*z)+{a1}*({b1}*z-{c1}*x)+{a1}*({b1}*x-{c1}*y)}=\simplify{{a1*b1-a1*c1}*x+{a1*b1-a1*c1}*y+{a1*b1-a1*c1}*z}.
b)
As in part a) the variables x, y, and z appear cyclically in each component of \boldsymbol{u}, so we only need calculate one derivative explicitly, then use cyclic permutations to calculate the other two. Hence
\begin{align}\frac{\partial u_1}{\partial x}&=\frac{\partial}{\partial x}\left\{\left(x^\var{p1}+y^\var{p1}\right)\left(\simplify{{a2}*y^{p1-1}-{a2}*z^{p1-1}}\right)\right\}\\&=\simplify{{p1}*x^{p1-1}}\left(\simplify{{a2}*y^{p1-1}-{a2}*z^{p1-1}}\right),\end{align}
and by symmetry
\frac{\partial u_2}{\partial y}=\simplify{{p1}*y^{p1-1}}\left(\simplify{{a2}*z^{p1-1}-{a2}*x^{p1-1}}\right),
and
\frac{\partial u_3}{\partial z}=\simplify{{p1}*z^{p1-1}}\left(\simplify{{a2}*x^{p1-1}-{a2}*y^{p1-1}}\right).
Finally, adding the derivatives together gives the divergence
\boldsymbol{\nabla\cdot u}=\simplify{{p1}*x^{p1-1}}\left(\simplify{{a2}*y^{p1-1}-{a2}*z^{p1-1}}\right)+\simplify{{p1}*y^{p1-1}}\left(\simplify{{a2}*z^{p1-1}-{a2}*x^{p1-1}}\right)+\simplify{{p1}*z^{p1-1}}\left(\simplify{{a2}*x^{p1-1}-{a2}*y^{p1-1}}\right)=0.
c)
First note that f_1 does not depend on z, f_2 does not depend on y, and f_3 does not depend on z. This makes the differentiation very straight forward, and hence
\begin{align}\boldsymbol{\nabla\cdot u}&=\frac{\partial}{\partial x}\left(\simplify{{a3}*x}+f_1(y,z)\right)+\frac{\partial}{\partial y}\left(\simplify{{b3}*y}+f_1(x,z)\right)+\frac{\partial}{\partial z}\left(\simplify{{c3}*z}+f_1(x,y)\right)\\&=\simplify[all,!collectNumbers]{{a3}+{b3}+{c3}}\\&=\var{a3+b3+c3}.\end{align}