O.K, here is a better answer; I think the other answer is also informative ( actually, I put in a lot of time into writing it , so I don't want to delete it :) ):
1) A tensor f in $\mathbb R^4$ is a function defined on $\mathbb R^4 $ that is linear on each variable separately, i.e., f is linear in each variable when other variables are held constant. In our case, the variables in $\mathbb R^4$ are $x=(x_1,x_2,..,x_4),...,z= (z_1, z_2, z_3, z_4)$. I think what is confusing is the way the function is expressed, as if it was a function defined on $\mathbb R^9$ , or so. I think this can be remedied by using the projection maps $\pi_j ; j=1,2,3,4$, with $\pi_j(x):=x_j$. Then the map $$ f(x,y,z)=3x_1y_2z_3 -x_3y_1z_4$$ , can be written in terms of $(x,y,z)$ alone (as f(x,y,z) ought to, since by definition it only depends on these 3 variables ) by $$f(x,y,z) = 3\pi_1(x)\pi_2(y)\pi_3(z)-\pi_3(x)\pi_1(y)\pi_4(z)$$.
This is a (composite) function of the three variables $(x,y,z)$ , just like, say , $cosx^2+tany$ is a function of two variables.
We know check that this f is a tensor, i.e., that f is linear on each of the variables $(x,y,z)$ ( notice $f$ is defined on a subspace of $\mathbb R^4$ consisting of the first 3 axes.). So we need to check, for all constants $c,d,k$ :
i ) $f(cx,y,z)=cf(x,y,z); f(x,dy,z)=df(x,y,z); f(x,y, kz)=kf(x,y,z)$.
Let me just do the part for $x$; the rest will likely be clear (let's forget the issue with the projections for now ):
$$f(cx,y,z):=f((cx_1,cx_2, cx_3, cx_4),(y_1,..,y_4),(z_1,..,z_4)=3(c_1)y_2 z_3-(cx_3)y_1z_4=c(x_1y_2z_3-x_3y_1z_4)=cf(x,y,z)$$
Now we need to check, for $x'=(x_1',x_2', x_3', x_4')$:
ii)$$ f(x+x',y,z)=f(x,y,z)+f(x',y,z) $$ (same for the $y,z $ variables). We have:
$$ f(x+x',y,z)=3(x_1+x_1')y_2z_3-(x_3+x_3')y_1z_4=3x_1y_2z_3+ 3x_1'y_2z_3 -(x_3y_1z_4)-(x_3')y_1z_4 $$ , which you can verify equals $f(x,y,z)+f(x',y,z)$ . A similar argument shows $f$ is linear with respect to $y,z$.
Now, to verify the correctness of the representation: $$3w^1 \otimes w^2 \otimes w^3 -w^3 \otimes w^1 \otimes w^4 :$$ , using the definition $w^i(e_j)= \delta_i^j$ , where $e_i$ ; i=1,2,3,4 is the "standard" basis $e_1=(1,0,0,0),..,e_4=(0,0,0,1)$.
In this basis, a point $x=(x_1,x_2,x_3, x_4)$ has a representation $$x_1e_1+x_2e_2+x_3e_3+x_4e_4 $$ . Also, the tensor operation $w^i \otimes w^j \otimes w^k$ is defined by $$w^i \otimes w^j \otimes w^k (a,b,c):=w^i(a)w^j(b)w^k(c) $$ , where $a,b,c$ are vectors.
We then have :
$$ 3w^1 \otimes w^2 \otimes w^3 ((x_1e_1+...x_4e_4),...,(z_1e_1,...,z_4e_4)=3w^1(x_1e_1,..x_4e_4)\otimes w^2(y_1e_1,..,y_4e_4) \otimes w^3(z_1e_1,..,z_4e_4)=3x_1y_2z_3 $$ . Try verifying the other component $w^3 \otimes w^1 \otimes w^4 $.
Hope that was helpful.