If I have $\psi(x)$ be a function such that $\psi = exp(g)$ with $g\geq 0$ on a bounded domain $X$ and for $u\in C_c^{\infty}(X)$ and a constant $c_1$, we have $A=-\Delta -c_{1}^2|\nabla \psi|^2$ and $Bu=c_1(\nabla\cdot(u\nabla \psi)+\nabla\psi\cdot \nabla u)$.
Show that $B$ is antysymmetric and expand the commutator $[A,B]=AB-BA$
I really don't understand how prove that $B$ is antisymmetric, but my idea was show that $B=-B^t$, so we have $$ (B)^tu=c_1(\nabla\cdot(u\nabla \psi)+\nabla\psi\cdot \nabla u)^t $$ But I don't see how take minus sign. And for the other point,
$$ [A,B]=[-\Delta -c_{1}^2|\nabla \psi|^2][c_1(\nabla\cdot(u\nabla \psi)+\nabla\psi\cdot \nabla u)]-[c_1(\nabla\cdot(u\nabla \psi)+\nabla\psi\cdot \nabla u)][-\Delta -c_{1}^2|\nabla \psi|^2] $$
Someone can give me some hint, thanks!!
Following the notation (and my comment) of your previous question, lets relabel $P_f=e^{\tau f}(-\Delta)e^{-\tau f}$ and recall that \begin{align*} P_fu&=e^{\tau f}(-\Delta)e^{-\tau f}u\\ &=-\Delta u-\tau^2 u|\nabla f|^2+\tau\nabla( u\nabla f)+\tau \nabla u\nabla f\\ &=A+B. \end{align*} So, now observe that \begin{align*} P_f=\dfrac{1}{2}(P_f+P_f)=\dfrac{1}{2}(P_f+P_{-f})+\dfrac{1}{2}(P_f-P_{-f}) \end{align*} which gives u a decomposition of $P_f$ in a symmetric part and antisymmetric part in terms of the adjoint operator. Notice that $P_{-f}=\Delta u+\tau^2 u|\nabla f|^2-2\tau\nabla u\nabla f-\tau u\nabla f$ and so trivially, \begin{align*} A=\dfrac{1}{2}(P_f+P_{-f})=-\Delta u-\tau^2 u|\nabla f|^2 \end{align*} and \begin{align*} B=\dfrac{1}{2}(P_f-P_{-f})=\tau\nabla( u\nabla f)+\tau \nabla u\nabla f. \end{align*} So it remains to check that for each $u$, $v\in C_c^\infty(X)$ it holds \begin{align*} (Au, v)_{L^2}=(u, Av)_{L^2}\ \text{ and }\ (Bu, v)_{L^2}=(u, -Bv)_{L^2} \end{align*} which follows directly performing integration by parts. The idea behind is that the Laplacian $\Delta$ and the multiplication operator $\tau^2|\nabla f|^2$ are self-adjoint in $H_0^1$ (operators involved in $A$) but the taking gradient $\nabla$ would be skew-adjoint in $H_0^1$ (the operator involved in $B$).
Again, as I stated in your other question, for the computation of the commutator $[A, B]$ you only need to follow the corresponding differentiabiliy rules in several variables (heavily used in any vector calculus course), you need to convince yourself there. As an advice, you can compute it by taking the corresponding partial derivatives and then adding up all the terms (many indices will appear of course). You should be also aware of how do you get the Hessian by these rules and how the product of matrices (Hessians) would look like.