$\newcommand{\id}{\operatorname{Id}}$ $\newcommand{\TM}{\operatorname{TM}}$ $\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{\Cof}{\operatorname{Cof}}$ $\newcommand{\Det}{\operatorname{Det}}$ $\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\TM}{\operatorname{T\M}}$ $\newcommand{\TN}{\operatorname{T\N}}$ $\newcommand{\TstarM}{T^*\M}$

Let $\M,\N$ be $d$-dimensional oriented Riemannian manifolds, and let $f:\M \to \N$ be smooth; Note that $df \in \Omega^1(\M,f^*{\TN}),\bigwedge^2 df\in \Omega^2\big(\M,\Lambda_2(f^*{\TN})\big)$. Let $\nabla^{\Lambda_2(f^*{\TN})}$ be the induced connection on $\Lambda_2(f^*{\TN})$ and let $\delta_{\nabla^{\Lambda_2(f^*{\TN})}} $ be the adjoint of $d_{\nabla^{\Lambda_2(f^*{\TN})}}$.

Consider the functional $ E(f):=\frac{1}{2}\int_{\M} \| \bigwedge^2 df\|^2 \text{Vol}_{\M}.$

Does there exist a critical point of $E$, $f:\M \to \N$, which does not satisfy $\delta_{\nabla^{\Lambda_2(f^*{\TN})}} \big( \bigwedge^2 df \big) =0$?

I proved here that the Euler-Lagrange equation of $E$ is $$h_{f^*\TN} \big( \tr_{\TM}\big( df \otimes \delta_{\nabla^{\Lambda_2(f^*{\TN})}}(\bigwedge^2 df)\big) \bigg)=0,$$

where $h_{f^*\TN}:f^*\TN \otimes \Lambda_2(f^*\TN) \to f^*\TN$ is the linear extension of

$$ \tilde w \otimes (w_1 \wedge w_2) \to \langle \tilde w,w_2 \rangle w_1-\langle \tilde w,w_1 \rangle w_2.$$

Edit:

The answer is negative for $d=2$, so we need to restrict our attention to $d \ge 3$.

Indeed, in this case $E(f)=\int_{\M}\| \bigwedge^2 df\|^2 \text{Vol}_{\M}=\int_{\M}(\Det df)^2 \text{Vol}_{\M}$. We shall prove a map $f$ is $E$-critical if and only if its determinant is constant.

Since $\Det df$ is constant $\iff$ $\nabla (\bigwedge^2 df)=0 \iff \delta_{\nabla^{\Lambda_2(f^*\TN)}} \big( \bigwedge^2 df \big)=0$, we are done.

The Euler-Lagrange equation of $E$ can be written as

$$ \delta (\Det df \cdot \Cof df) =0, \tag{1}$$ where $\Cof df:\TM \to f^*(\TN)$ is the cofactor map of $df$ defined by $$ \Cof df= (-1)^{d-1} \star_{f^*TN}^{d-1} (\wedge^{d-1} df) \star_{TM}^1, $$

and $\delta$ is the adjoint of the pullback connection on $f^*\TN$.

(The Euler-Lagrange equation of the Jacobian functional $E(f)=\int_{\M} \Det df \text{Vol}_{\M}=\int_{\M} f^* \text{Vol}_{\N}$ is $\delta (\Cof df)=0$. Details can be found in lemma 2.9 in my paper here).

Expanding equation $(1)$ we get $$ 0=\delta(\Det df \cdot \Cof df )= \Det df \delta(\Cof df ) - \tr_{g}(d \Det df \otimes \Cof df). $$

We now use the fact the Jacobian functional is a null-Lagrangian, i.e. every smooth map is a critical point, or equivalently $\delta (\Cof df)=0$. (This is essentially Stokes theorem, you can see lemma 2.5 here).

So, the E-L equation $(1)$ reduces to $$\tr_{g}(d \Det df \otimes \Cof df)=0. \tag{2}$$

Let $f$ be a map satisfying equation $(2)$. We shall prove $\Det df$ is constant; suppose that $\Det df_p \neq 0$ for some $p \in \M$. This implies $ \Cof (df_p)$ is invertible, so by equation $(2)$ $d\Det df_p=0$.

Now we observe that any $C^1$ function $g : \M \to \mathbb R$ on a connected manifold, satisfying $g(p) \ne 0 \implies dg_p = 0$ is constant.