Problem 6 chapter 3 from Evans PDE 2nd edition I am working on the following problem

I have solved (a) but I'm struggling with (b). This is what I've done so far: First I modified the given equation:
$$u_t +div(u\mathbb{b})=u_t+Du\cdot\mathbb{b}+u\,div(\mathbb{b})=0.$$
Then I tried to solve it with the method of characteristics and got the following ODEs
$$\dot{\mathbb{x}}=\mathbb{b},\quad \dot z=-div (\mathbb{b})\, z,\quad \dot t=1$$
with initial conditions
$$\mathbb{x}(0)=a,\quad z(0)=g(a),\quad t(0)=0.$$
From these I then solve $z$ and get
$$z=g(a)e^{-div(\mathbb{b})}.$$
I don't know how to proceed (or if I've made any mistakes) and I have no idea how to use the hint or part (a). Any help with this one?
EDIT: I think my solution $z$ is incorrect, I didn't think about the fact that $\mathbb{b}$ depends on $\mathbb{x}$. I haven't yet figured out the correct one.
 A: (a) Try to use Jacobi identity: $$\frac{d}{ds}\det A(s)=\text{tr}\left((\text{cof}A(s))\frac{dA}{ds}(s)\right)$$
Let $A(s)=\left(X_{x_j}^{i}(s,x,t)\right)$ and $B(s)=\left(b_{x_j}^{i}(s)\right)$. We yield $$\frac{dA}{ds}(s)=B(X)A(s)$$
Plug in the identity, done.
(b) Using method of characteristics is fine, $\dot{x}(t)=b$, $x(0)=y$, $\dot{z}(t)=-\text{div}bz^{-1}$, $z^{-1}(0)=g^{-1}(y)$.
Note that by assumption of (b) and Jacobi identity in (a), these ODEs have unique solutions. Moreover,
$$y=X(-t,x,0), X\left(s,X(-t,x,0),0\right)=X\left(X(s-t,x,0),0\right)$$
$$J(-t,x,0)=J(0,x,t)$$
$$z(t)=\frac{g(y)}{J(t,y,0)}$$
By Euler formula,
$$J(t,y,0)=\exp\left(\int_{0}^{t}\text{div}b(X(s,y,0))ds\right)=\exp\left(\text{div}\left(X(s,X(-t,x,0),0)\right)ds\right)=\exp\left(\int_{0}^{t}\text{div}b(X(s-t,x,0))ds\right)=\exp\left(-\int_{-t}^{0}\text{div}b(X(\tau,x,0))d\tau\right)=J^{-1}(-t,x,0)=J^{-1}(0,x,t)$$
This suggests $$u(x,t)=g(X(0,x,t))J(0,x,t)$$
A: a) Using the property $\partial\Delta/\partial {\bf A} = \Delta{\bf A}^{-\text{T}}$ of the determinant $\Delta = \det {\bf A}$, the chain rule gives Euler's formula
\begin{aligned}
J_s &= J (D_x {\bf x})^{-\text{T}} \cdot \partial_s D_x {\bf x} \\ 
&= J (D_x {\bf x})^{-\text{T}} \cdot D_x \dot{\bf x} \\
&= J (D_x {\bf x})^{-\text{T}}\cdot D_x {\bf b}({\bf x}) \\
&= J \operatorname{tr}( (D_x {\bf x})^{-1} D_x {\bf b}({\bf x}) ) \\
&= J \operatorname{tr}( D_{\bf x} x\, D_x {\bf b}({\bf x}) ) \\
&= J \operatorname{div} {\bf b}({\bf x})
\end{aligned}
where the dot $\cdot$ denotes the Frobenius inner product (scalar product of matrices).
b) Here, we only need to verify that the proposed solution solves the PDE problem. Firstly, one notes that at $t=0$,
\begin{aligned}
u(x,0) = g({\bf x}(0,x,0))J(0,x,0) = g(x) .
\end{aligned}
This follows from evaluation of the solution to the previous boundary-value problem for $\dot{\bf x} = {\bf b}({\bf x})$ at $s=t=0$.
Secondly, one notes that if the hint property is true (which shouldn't be too hard to prove), then for all $s$, we have
$$
u({\bf x}, s)J = u(x,t) := g({\bf x}(0,x,t)) J(0,x,t)
$$
by evaluation of $u({\bf x}, s)J$ at $s=t$. It remains to prove that the proposed expression satisfies the PDE at hand. By using the differentiation rules in the hint, we find
$$
0 = J u_{x}({\bf x}, s)\cdot {\bf b} + J u_t({\bf x}, s) + J_s u({\bf x}, s)
$$
which provides the PDE $u_t + \operatorname{div} (u{\bf b}) = 0$ by using a).
