Let $(x,y)$ evolve to $(X,Y)$ under the velocity field $\vec v=(u,v)$ in time $t$. Thus, $$X=X(x,y,t), Y=Y(x,y,t), t=t$$ Then, $$J(x,y,t)= \begin{bmatrix} \frac{\partial X}{\partial x} & \frac{\partial X}{\partial y} & \frac{\partial X}{\partial t} \\[1ex] % <-- 1ex more space between rows of matrix \frac{\partial Y}{\partial x} & \frac{\partial Y}{\partial y} & \frac{\partial Y}{\partial t} \\[1ex] \frac{\partial t}{\partial x} & \frac{\partial t}{\partial y} & \frac{\partial t}{\partial t} \end{bmatrix}$$ $$=\begin{bmatrix} \frac{\partial X}{\partial x} & \frac{\partial X}{\partial y}\\[1ex] \frac{\partial Y}{\partial x} & \frac{\partial Y}{\partial y} \end{bmatrix}$$ Now $\frac{\partial J}{\partial t}=\frac{\partial u}{\partial x}\frac{\partial Y}{\partial y}+\frac{\partial X}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial u}{\partial y}\frac{\partial Y}{\partial x}-\frac{\partial X}{\partial y}\frac{\partial v}{\partial x}$.
Thus, $\frac{\partial J}{\partial t}$ is the trace of matrix $\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\[1ex] \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{bmatrix}.\begin{bmatrix} \frac{\partial Y}{\partial y} & -\frac{\partial X}{\partial y}\\[1ex] -\frac{\partial Y}{\partial x} & \frac{\partial X}{\partial x} \end{bmatrix}=\begin{bmatrix} \frac{\partial u}{\partial X} & \frac{\partial u}{\partial Y}\\[1ex] \frac{\partial v}{\partial X} & \frac{\partial v}{\partial Y} \end{bmatrix}.J(x,y,t)$ Therefore, $\frac{\partial J}{\partial t}=(\nabla.\vec v)J$.
Now how did the steps after "trace of matrix" conclude? Also, why are we taking gradient with respect to $X,Y$ and not $x,y$ in the last step?