The normal density looks like $f(x) \propto \exp\left( -\frac{(x-\mu)^2}{2\sigma^2} \right)$.
Suppose that you know that $X \sim N(\cdot, \cdot)$ is normal, but the density looks like $f_X(x) \propto \exp\left( g(x) \right)$, i.e. it isn't clear by inspection what $\mu$ and $\sigma^2$ are supposed to be. We know by symmetry that $\mathbb{E}(X)$ occurs at the peak of the density, so we can get the mean by differentiating the exponent $g(x)$ w.r.t. $x$ and setting to 0. My question is: is there a similar trick to find the variance of $X$ when $f$ is not in the standard form? What is the fastest way to find the variance of $X$ in this situation?
The normal distribution has inflection points at plus and minus one standard deviation from its mean, i.e. at $x=\mu+\sigma$ and $x=\mu-\sigma$, so differentiate $f(x|\mu,\sigma^2)$ twice and set equal to $0$, finding the points of inflection, then take half the difference between them for the standard deviation. Finally, square for the variance.
You could also attempt to complete the square.