$\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\R}{\mathbb{R}}$
Does there exist a smooth map $f:\mathbb{R}^3 \to \mathbb{R}^3$, which satisfies $$\tr \big( df \otimes \delta(df \wedge df) \big)=0, $$ such that $\delta \big( df \wedge df \big) $ does not vanish identically?
Here $\bigwedge^2 df\in \Omega^2\big(\R^3,\Lambda_2(\R^3)\big)$, and $ \delta \big( df \wedge df \big)\in \Omega^1\big(\R^3;\Lambda_2(\R^3)\big)$, where $\delta$ is the adjoint of the exterior derivative.
If there is such an $f$ then it cannot be a local diffeomorphism; its Jacobian must vanish at some point. Expanding explicitly
$$ \delta \big( df \wedge df \big)(X)=\sum_i \nabla_{e_i}(df \wedge df)(e_i,X)=\sum_i (\nabla_{e_i} df)(e_i) \wedge df(X) + df(e_i) \wedge (\nabla_{e_i} df)(X)$$ $$ =\Delta f \wedge df(X)+\sum_i df(e_i) \wedge (\nabla_{e_i} df)(X).$$
Thus $$ \delta \big( df \wedge df \big)=\Delta f \wedge df+\sum_i df(e_i) \wedge \nabla_{e_i} df. $$ We can try to look first for harmonic counter-examples, but I failed doing even that.
Motivation: This question is a degenerate version of this unsolved question about critical points.
