How to find the maximum curvature of $y=e^x$? So I found the curvature to be $K = \dfrac{e^{x}}{(1+e^{2x})^{3/2}}$ but I don't know how to maximize this?

We can differentiate and set equal to zero to get
$$\frac{e^x-2e^{3x}}{(1+e^{2x})^{5/2}}=0\implies e^x=2e^{3x}\implies\frac{1}{2}=e^{2x}.\tag{1}$$
Then solve to obtain $x=-\frac{1}{2}\log 2$.
Note - the various algebraic operations in (1) are possible since $e^x\neq 0$ for all $x\in\mathbb{R}$. For future similar questions, it may be necessary to determine maxima/minima using the second derivative. So called "saddle" points may also need addressing.
