Let $A \in M_{n, n}(\mathbb{R})$ be a symmetric matrix, and let $N = \{x \in \mathbb{R}^n \mid x^t A x = 0\}$.
I first want to show that $N$ is closed. Next, I want to find an explanation why $f(x) = \frac{1}{x^t A x}$ is differentiable on $\mathbb{R}^n\setminus N$, and I want to find it's derivative.
To the first problem, my attempt is: $g: \mathbb{R}^n \to \mathbb{R}, x \mapsto x^t A x$ is continuous because it's a bilinear form. So if $(x_n) \subseteq \mathbb{R}^n \to x_0$ for any $x_0 \in \mathbb{R}^n$ with $g(x_n) \in N \Longleftrightarrow g(x_n) = 0$ for all $n$, then $g(x_0)$ must be $0$ too, because otherwise we get a contradiction to the continuity of $g$, and therefore, $N$ is closed. Would this be sufficient?
To the second problem, I don't really know what to do though. We know $f$ is continuous, but how do we get differentiability?