$f(\alpha,\beta)=g(\alpha)h(\beta)$, where $f$ is symmetric bilinear, $g,h$ are linear. 
$f(\alpha,\beta)=g(\alpha)h(\beta)$, where $f$ is symmetric bilinear on a vector space $V$, $g,h$ are linear function on $V$. show $f(\alpha,\beta)=\lambda k(\alpha)k(\beta)$ for a number $\lambda$ and linear function $k$ on $V$.

Clearly, if $f=0$, it is ok. If $f\neq 0$, $g(\alpha)h(\beta)=f(\alpha,\beta)=f(\beta,\alpha)=g(\beta)h(\alpha)$. How can we divide $g(\beta)h(\beta)$? we do not know it is $\neq 0$.

How to prove the above statement?

 A: Let's assume that $f \not = 0$. Let's assume now that still, there exists an $\alpha \in V$ such that $f(\alpha,\alpha) = 0$. This implies that $g(\alpha)=0$ or $h(\alpha)=0$. W.l.o.g. assume it is $g(\alpha)=0$. Then for any $x \in V$ we'll have
$$f(\alpha,x)=g(\alpha)h(x) = 0 = f(x,\alpha) = g(x)h(\alpha)$$
From which we can conclude that either $h(\alpha) = 0$, either that $g(x)=0$. But the latter would imply that it is true for any $x \in V$, in contradiction with our initial assumption that $f \not = 0$. So, we have to accept that if for some $\alpha \in V$, $g(\alpha)=0 \implies h(\alpha)=0$ and visa versa of course.
Hence, we can in your formula divide by the product $g(\beta)h(\beta)$, at least for some $\beta \in V$. Let's just call it $\gamma$ from now on. From which we learn that
$$\exists \gamma \in V: \forall \alpha \in V: \frac{g(\alpha)}{g(\gamma)}=\frac{h(\alpha)}{h(\gamma)}$$
Let's relabel that special
Thus,
$$f(\alpha,\beta) = g(\alpha)h(\beta) = g(\alpha)g(\beta)\frac{h(\gamma)}{g(\gamma)}$$
So, with $\lambda = \frac{h(\gamma)}{g(\gamma)}$ and $k = g$, we obtain the result.
