The trick is the following:
$$ \epsilon_{ijk} \frac{\partial u_m}{\partial x_j} \frac{\partial u_m}{\partial x_k} = 0 $$
by antisymmetry.
So you can rewrite
$$ \epsilon_{ijk} \frac{\partial u_m}{\partial x_j} \frac{\partial u_k}{\partial x_m} = \epsilon_{ijk} \frac{\partial u_m}{\partial x_j}\left( \frac{\partial u_k}{\partial x_m} - \frac{\partial u_m}{\partial x_k} \right) $$
Note that the term in the parentheses is something like $\pm\epsilon_{kml} \omega_l$
Lastly use the product property for Levi-Civita symbols
$$ \epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} $$
Linear Independent Case
$1)$ If $A$, $B$, and $C$ are three independent vectors then $A \times B$, $B \times C$, and $C \times A$ are also independent so they can form a basis for $\mathbb{R}^3$. So we can write any vector as a linear combination of them.
$2)$ According to step $(1)$ we have
$$(A \cdot (B \times C))D = \alpha (A \times B) + \beta (B \times C) + \gamma (C \times A)$$
where $\alpha$, $\beta$, and $\gamma$ are unknown coefficients. If we dot product with $A$, $B$, and $C$, respectively, we can get
$$\eqalign{
& \left( {A \cdot (B \times C)} \right)\left( {A\cdot D} \right) = \beta A\cdot (B \times C) \cr
& \left( {A \cdot (B \times C)} \right)\left( {B \cdot D} \right) = \gamma B \cdot (C \times A) \cr
& \left( {A \cdot (B \times C)} \right)\left( {C \cdot D} \right) = \alpha C \cdot (A \times B) \cr} $$
$3)$ We note that
$$A \cdot (B \times C) = B \cdot (C \times A) = C \cdot (A \times B)$$
$4)$ We finally conclude that
$$\eqalign{
& \alpha = C \cdot D \cr
& \beta = A \cdot D \cr
& \gamma = B \cdot D \cr} $$
Linear Dependent Case
Consider the case when the vectors $A$, $B$, and $C$ are linearly dependent and hence they cannot form a basis for $\mathbb{R}^3$. This means that there exists real numbers $a$ and $b$ not both zero such that
$A = aB + bC$
will hold. This will lead to $(A \cdot (B \times C))=0$ and turns our equation into a trivial identity $0=0$.
Best Answer
The first and third Levi-Civita symbols contract to $\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$, so the component is$$\color{blue}{\epsilon_{pqj}}(A_pB_qC_iD_j-A_pB_qC_jD_i)=(A\cdot B\times D)C_i-(A\cdot B\times C)D_i,$$where the blue factor uses $\epsilon_{jpq}=\epsilon_{pqj}$.
A proof with less calculation uses the fact that, since $C\times D$ is orthogonal to the plane $C,\,D$ span, $(A\times B)\times(C\times D)$ is a linear combination of $C,\,D$ that changes sign when they're exchanged, and also when $A,\,B$ are. This, together with multilinearity, gives the desired result up to a proportionality constant, which can be fixed with an example.