Let $f: V \times V \to F $ be a bilinear form and $f$ has the follow property: Whenever $f(u,v)=0$ we have $f(v,u)=0$. The goal is to prove that $f$ is symmetric or $f$ is skew - symmetric, for that prove the follow statements.
$a)$ Let's $u,v,w \in V$, prove that $f(u, f(u,w)v - f(u,v)w) =0$. Then show that $f(u,w)f(v,u)= f(w,u)f(u,v)$.
$b)$ Now prove that $w=u$ implies $f(u,u)=0$ if $f(u,v) \neq f(v,u)$.
$c)$ Show that if $f$ is now symmetric then $f(w,w)=0, \forall w \in V$. For that, choose $u,v$ s.t $f(u,v) \neq f(v,u)$ then observe the cases $f(u,w) \neq f(w,u)$ and $f(u,w) \neq f(w,u)$
$d)$Show that if $f(w,w)=0 \forall w \in V$ then $f$ is skew symmetric.
$\textbf{My attempt:}$
a) I don't know how to prove that $f(u, f(u,w)v - f(u,v)w) =0$, but if that was true then it is easy to show that $f(u,w)f(v,u)= f(w,u)f(u,v)$.
b) if $f(u,w)f(v,u)= f(w,u)f(u,v)$ and $w=u$ it is easy to show that $f(u,u)=0$ by $f(u,v) \neq f(v,u)$.
The other itens I don't know how to continue, can you help me?
Hint
for the very first one, you want to prove $$ f(u, f(u,w)v - f(u,v)w) =0 \tag{1} $$
All you know about $f$ is that it's bilinear and has the special property. The special property doesn't look all that relevant here, so what happens if you expand the left-hand side of equation 1 using bilinearity?
Also: in working on problems like this, it's good to have an example in mind. In this case, I recommend thinking about $V = \Bbb R^2$, and $f(u, v) = u \cdot v$. Then it's easy to see that the thing your attempted solution was based on can't possibly be true in general, so that's a blind lead.