Lie bracket of exact differential one-forms

Answer: The answer to my question is that $[dA,dB]$ is not necessarily zero. As a counterexample, consider $\mathbb{R}^2$ with the standard inner product and take $A(x,y) = x^2y^2$, $B(x,y) = xy^3$ (I selected these arbitrarily without thinking about them). Then $\nabla A(x,y) = (dA)^\sharp(x,y) = 2xy^2\frac{\partial}{\partial x} + 2x^2y \frac{\partial}{\partial y}$ and $\nabla B(x,y) = (dB)^\sharp(x,y) = y^3 \frac{\partial}{\partial x} + 3xy^2\frac{\partial}{\partial y}$.

Computing the Lie Bracket of $\nabla A$ and $\nabla B$ in coordinates shows that $[dA,dB](x,y) = [\nabla A, \nabla B](x,y) = (-6x^2y^3 - 2y^5)\frac{\partial}{\partial x} + (2xy^4 + 6x^3y^2)\frac{\partial}{\partial y}$ which is not zero at all points.

Resolving my misunderstanding: The following is an explanation of the misunderstanding which prompted my question. Let $A_1,\ldots,A_k$ be functions on a Riemannian $k$-manifold $(M,g)$ such that the collection of vectors $\{\nabla A_i := (dA_i)^\sharp\}$ are linearly independent at the point $x_0 \in M$. This is true if and only if the collection of forms $\{dA_1,\ldots,dA_k\}$ are themselves linearly independent as linear functionals at $x_0 \in M$. Then the derivative of the map $\varphi: M \to \mathbb{R}^k$, $\varphi: x \mapsto (A_1(x),\ldots,A_k(x))$ is full rank at $x_0 \in M$, so the inverse function theorem guarantees that there exists an open set $U \ni x$ such that $\varphi|_U: U \to \varphi(U)$ is a diffeomorphism. Initially, I thought that the pairwise Lie brackets of all of the $\nabla A_i$ must therefore be zero. However, this is only necessarily be true if the integral curves of the $\nabla A_i$ were the coordinate lines determined by the chart $\varphi$. The definition of $\nabla A_i$ relies on the particular metric $g$, and in general $g$ has nothing to do with $\varphi$; the coordinate lines of $\varphi$ which have nothing do with $g$ are thus not the integral curves of the $\nabla A_i$ in general.