I feel like I'm messing something up badly in the proof but I'm not able to see it, so I'm writing the proof down in hopes that someone catches an error:
Let $f$ be a $3$ linear function, $f: V^3 \rightarrow \mathbb{R}$
Define the alternating operator $A$, which converts any $3$-linear map into an alternating $3$-linear map.
$$ A : (V^3 \rightarrow \mathbb{R}) \rightarrow (V^3 \rightarrow \mathbb {R}) \\ A(f) = \sum_{\sigma \in S_k} \text{sgn}(\sigma) (\sigma f) $$ where $S_k$ is the symmetric group on $k$ letters, and $\sigma f$ is the action of $\sigma$ on the inputs of $f$. That is, $$(\sigma f) (x_1, x_2, x_3) \equiv f(x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}) $$
We are now asked to compute $A(f)$ for any $f : V^3 \rightarrow \mathbb{R}$, $f$ multilinear.
Proof
Let us evaluate $A(f)$ on some $(x, y, z)$.
\begin{align} A(&f)(x, y, z) = \\ &f(x, y, z) + \\ -&f (x, z, y) + \\ -&f(z, y, x) + \\ -&f(y, x, z) + \\ &f(z, y, x) + \\ &f(y, z, x) \end{align}
Which on using the linearity of $f$ yields: $$A(f) = f(0, 0, 0)$$
(Notice that the signs on the $x$ and $-x$ cancel out, as do for $y$ and $z$)
Is this proof correct? If so, what is the intuition for this? Why is only the "degenerate" solution popping up? Is it something to do with the odd dimension of the vector space? (dim = 3)
The linearity holds for one entry only when other entries are unchanged. $$A(f)(x, y, z) = f(x, y, z) + f(y, z, x) + f(z, x, y) - f(x, z, y) - f(y, x, z) -f(z, y, x)$$
None of these will cancel each other if we only assume $f$ is 3-linear. However, if we assume f is symmetric we will get 0.