Let $V$ be a real vector space of dimension $2$, and let $\langle\ \ ,\ \ \rangle$ be an inner product on $V$. Define $f:V^4 \to \mathbb{R}$ by $$f(x,y,z,w):=\langle x,y \rangle \langle z,w \rangle- \langle x,w \rangle \langle z,y \rangle$$
Show there's a skew-symmetric bilinear form $g:V^2 \to \mathbb{R}$ such that $$f(x,y,z,w)=g(x,z)g(y,w)$$
My thought: Let $z=y,w=x$ to get $f(x,y,y,x)= \langle x,y \rangle ^2-\langle x,x \rangle \langle y,y \rangle$. If $g$ exists, then $f(x,y,y,x)=g(x,y)g(y,x)=-g(x,y)^2$ (notice that $g$ is skew-symetric). Equating both identities we get $g(x,y)^2= -\langle x,y \rangle ^2+\langle x,x \rangle \langle y,y \rangle$. Here is where I can't proceed any more...