Derivation of the shape of the catenary curve, clarification of a particular step
If you allow $z(x)=y'(x)$, then (5) is $$ az'(x) = \sqrt{1+z^2}. $$
By separating variables: $$ \frac{dz}{\sqrt{1+z^2}} = \frac{dx}{a}, $$
which can be integrated to $$ \mathop{\mathrm{arcsinh}} z = \frac{x-x_0}a. $$