$\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\R}{\mathbb{R}}$
I am looking for a smooth map $f:\mathbb{R}^3 \to \mathbb{R}^3$, which satisfy
$$\delta \big( df \wedge df \big) \neq 0, \, \text{and } \, \tr \big( df \otimes \delta(df \wedge df) \big)=0, $$
where $ \delta \big( df \wedge df \big)\in \Omega^1\big(\R^3;\Lambda_2(\R^3)\big)$ is a one-form on $\R^3$ with values in $\Lambda_2(\R^3)$, defined by
$$ \delta \big( df \wedge df \big)(X)=\sum_i \nabla_{e_i}(df \wedge df)(e_i,X)=\sum_i (\nabla_{e_i} df)(e_i) \wedge df(X) + df(e_i) \wedge (\nabla_{e_i} df)(X), \tag{1}$$ for any vector field $X$. (Actually this is a tensor, defined pointwise, so we only need to know $X$ at a given point for evaluation). Here $\nabla_{e_i}$ is the standard component-wise derivative w.r.t the variable $x_i$, or the Levi-Civita connection of $\R^3$, if you prefer.
By the second equation we mean $$ \tr \big( df \otimes \delta(df \wedge df) \big):=\sum_i df(e_i) \otimes \bigg( \delta \big( df \wedge df \big)(e_i) \bigg)=0.$$
This is an equation in $\R^3 \otimes \Lambda_2(\R^3)$.
Simplifying equation $(1)$, we get
$$ \delta \big( df \wedge df \big)(X)=\Delta f \wedge df(X)+\sum_i df(e_i) \wedge (\nabla_{e_i} df)(X).$$
I know that if there is such a map $f$ then it cannot be a local diffeomorpihsm; its Jacobian must vanish somewhere.
Edit:
Clearly, $f$ must have some non-zero hessian; if $\nabla df=0$ then $\delta(df \wedge df)=0$. I tried putting quadratic polynomials but so far got nowhere.
Motivation: $\delta$ is actually the (minus of the) adjoint of the covariant derivative. This question is a special, degenerate case of the following general question:
Does every critical map of $E(f)=\int_{M}| df \wedge df|^2 \text{Vol}_{M}$ satisfy $\delta (df \wedge df) =0$? You can see more details here.