The differential proper time $d\tau$ is defined by
$$ c^2 d\tau^2 = -g_{\mu\nu}dx^\mu \otimes dx^\nu $$
So i believe we can think of $d\tau^2$ as $$ d\tau^2 = -\frac{1}{c^2}ds^2 $$ where $ds^2$ is the inner product.
I've read that \begin{equation} g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-c^2 \end{equation} How can I (rigorously) derive this last equation from the first? Is there a way? Physics texts say it is, but I have not seen it done (apart from obviously not acceptable "proofs" that simply "divide" the first equation by $d\tau^2$ on both sides).
Proper time for a timelike path $x^\mu(\lambda)$ is defined by $$ \tau = \frac{1}{c} \int \left( - g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}\right)^{1/2} d\lambda $$ where we integrate over a part of the path. So differentiating with respect to $\lambda$ and squaring we find $$ c^2 \left(\frac{d\tau}{d\lambda} \right)^2 = - g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}. $$
Using the chain rule we find $$ g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} = g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} \left(\frac{d\tau}{d\lambda} \right)^2, $$ which gives the required equation.
In fact you have probably already seen a very similar thing in differential geometry for regular curves in $\mathbb{R}^3$ parameterized by their arc length. Given an interval $I$, a curve $\alpha: I \to \mathbb{R}^3$ the arc length is defined by $$ s(t) = \int_{t_0}^t \lvert \alpha'(s) \rvert ds = \int_{t_0}^t \left( \alpha'(s) \cdot \alpha'(s) \right)^{1/2} ds. $$ Then we have $\frac{ds}{dt} = \lvert \alpha'(s) \rvert$ which is equal to $1$ if $\alpha$ is parameterized by its arc length.