Let $q_1$ be a definite positive quadratic form of $\mathbb{R}^n$ and $q_2$ be a quadratic form of $\mathbb{R}^n$. We define
$$ f(x) := \frac{q_2(x)}{q_1(x)}. $$
I want to prove that
$$ df_{|x} = \frac{2\varphi_2(x, \cdot)q_1(x) - 2\varphi_1(x, \cdot)q_2(x)}{q_1(x)^2}, $$
with $\varphi_1$ and $\varphi_2$ being the polar forms of $q_1$ and $q_2$ respectively. Thank you in advance for the help, I am not sure how differentiate this quotient.
Claim. If $f,g : \mathbb R^n \to \mathbb R$ are differentiable at $a \in \mathbb R^n$, and $g(a) \neq 0$, then $f/g : g^{-1}(\mathbb R \setminus \{0\}) \to \mathbb R$ is also differentiable at $a$ and its derivative $d(f/g)|_a : \mathbb R^n \to \mathbb R$ is the linear map $$d(f/g)|_a = \frac{g(a)df|_a - f(a)dg|_a}{[g(a)]^2} \ \bigg(\!\!= \frac1{g(a)} df|_a - \frac{f(a)}{[g(a)]^2} dg|_a \bigg).$$
Using the above result, it is sufficient show that if $b : \mathbb R^n \times \mathbb R^n \to \mathbb R$ is a symmetric bilinear map, and $q : \mathbb R^n \to \mathbb R$ is given by $q(x) = b(x,x)$, then $dq|_a =2 b(a,\cdot)$ for any $a \in \mathbb R^n$. Indeed, this is easy to show from the very definition of Fréchet derivative: $$h \in \mathbb R^n \setminus \{0\} \implies \frac{|q(a+h)-q(h)-2b(a,h)|}{\|h\|} = \frac{|b(a,a)|}{\|h\|}.$$