How do you find $\frac{d}{dx} \frac{x^T A x}{x^T B x}$? I only know that $\frac{d}{dx} x^T A x = 2A x$ if $A$ is symmetric. I saw the following on some notes but I don't know how I can reach this conclusion:
$$\frac{d}{dx} \frac{x^T A x}{x^T B x} = x^T A x\frac{d}{dx}(x^T B x) - x^T B x \frac{d}{dx} x^T A x$$
Your notes do not say that $$\frac{d}{dx} \frac{x^T A x}{x^T B x} = x^T A x\frac{d}{dx}(x^T B x) - x^T B x \frac{d}{dx} x^T A x.$$ Instead they say that $\frac{d}{dx} \frac{x^T A x}{x^T B x} = 0$ implies that $x^T A x\frac{d}{dx}(x^T B x) - x^T B x \frac{d}{dx} x^T A = 0$. This isn't the same thing, if $\frac{d}{dx} \frac{x^T A x}{x^T B x} \neq 0$ the right hand side will look very different.
To prove that $\frac{d}{dx} \frac{x^T A x}{x^T B x} = 0 \Rightarrow x^T A x\frac{d}{dx}(x^T B x) - x^T B x \frac{d}{dx} x^T A = 0$, note that $$ \frac{d}{dx} \frac{x^T A x}{x^T Bx} = \frac{1}{x^T Bx} \frac{d}{dx} (x^T A x) + (x^T Ax) \frac{d}{dx} \left( \frac{1}{x^T Bx} \right), $$ and $$ \frac{d}{dx} \left( \frac{1}{x^T Bx} \right) = - \frac{1}{(x^T B x )^2} \frac{d}{dx}(x^T B x). $$ So filling this in we have
$$ \frac{d}{dx} \frac{x^T A x}{x^T Bx} = \frac{1}{(x^T Bx)^2} \left[ (x^T B x) \frac{d}{dx} (x^T A x) - (x^T A x) \frac{d}{dx} (x^T B x) \right]. $$ So this confirms the formula in your question does not hold. But what is true is that if one side is zero, the other must be too.