Answer
The curvature at $s=0$ is $\kappa \left( 0 \right) = 1$.
Work Step by Step
We have $\left( {x\left( s \right),y\left( s \right)} \right) = \left( {\cosh s,s} \right)$.
The derivatives are
$\left( {x'\left( s \right),y'\left( s \right)} \right) = \left( {\sinh s,1} \right)$
$\left( {x{\rm{''}}\left( s \right),y{\rm{''}}\left( s \right)} \right) = \left( {\cosh s,0} \right)$
Using Eq. (11) we compute the curvature
$\kappa \left( s \right) = \frac{{\left| {x'\left( s \right)y{\rm{''}}\left( s \right) - x{\rm{''}}\left( s \right)y'\left( s \right)} \right|}}{{{{\left( {x'{{\left( s \right)}^2} + y'{{\left( s \right)}^2}} \right)}^{3/2}}}}$
$\kappa \left( s \right) = \frac{{\left| {0 - \cosh s} \right|}}{{{{\left( {{{\sinh }^2}s + 1} \right)}^{3/2}}}}$
Since ${\cosh ^2}s - {\sinh ^2}s = 1$, so
$\kappa \left( s \right) = \frac{{\left| {\cosh s} \right|}}{{{{\left( {{{\sinh }^2}s + 1} \right)}^{3/2}}}} = \frac{{\cosh s}}{{{{\cosh }^3}s}} = \frac{1}{{{{\cosh }^2}s}}$
The curvature at $s=0$ is $\kappa \left( 0 \right) = 1$.
