Proving $(\bf x\times y\cdot N)\ z+(y\times z\cdot N)\ x+(z\times x \cdot N)\ y= 0$ when $\bf x,y,z$ are coplanar and $\bf N$ is a unit normal vector

If what is required is only to prove the validity of the given identity, there is another approach. Observe that if $x$ and $y$ are linearly dependent, i.e. for some $c$, $x=c y$ or $y=c x$, then the identity holds trivially because $w\times v =-(v\times w)$ and $v \times v=0$ for all $v,w$. Thus, we may assume $x$ and $y$ are linearly independent and hence $z$ is a linear combination of $x$ and $y$, that is, $z=ax+by$ for some $a,b$. Now, since the given identity is linear in each variable and it holds for both $z=x$ and $z=y$, it is also true for $z=ax+by$. This proves the identity. It can be also noted that $N$ being perpendicular to the plane containing $x,y,z$ plays no role in this proof.

Here's an observation: If $Q$ is a rotation matrix, then $$ (Qx) \times (Qy) = Q(x \times y) $$

You have to prove that, of course, but it's not too tough. Similarly, $$ (Qx) \cdot (Qy) = x \cdot y $$ and, for a scalar $\alpha$, we have $$ Q (\alpha x) = \alpha (Q x) $$

Now suppose that for some vector $v$, we have $$ (\mathbf{x}\times\mathbf{y} \cdot \mathbf{N})\ \mathbf{z} + (\mathbf{y}\times\mathbf{z} \cdot \mathbf{N})\ \mathbf{x} + (\mathbf{z}\times\mathbf{x} \cdot \mathbf{N})\ \mathbf{y}=\mathbf{v}. $$

Key idea 1: You can apply the rules above to show that for any rotation matrix $Q$, you can apply $Q$ to all the elements on the left to get $Qv$.

Key idea 2: You can choose $Q$ so that it takes $N$ to the vector $(0,0,1)$, and puts $x, y,$ and $z$ into the plane consisting of vectors of the form $(a, b, 0)$. And in that plane, it's easy to see that you get $0$, so $Qv = 0$. Hence $v = 0$, and you're done.

In short: by a change of basis, you can assume that $N$ is the vector $(0,0,1)$ and that the other vectors all lie in the $(a, b, 0)$ plane, and things get easy.

Writing $x=a\hat{i}+b\hat{j},\,y=c\hat{i}+d\hat{j},\,z=e\hat{i}+f\hat{j},\,N=N\hat{k}$ reduces the sum to $$N((ad-bc)(e\hat{i}+f\hat{j})+(cf-de)(a\hat{i}+b\hat{j})+(be-af)(c\hat{i}+d\hat{j})).$$The $\hat{i}$ coefficient is $N(ade-bce+acf-ade+bce-acf)=0$. The $\hat{j}$ coefficient can be handled similarly.