The following is part of the solution of an Exercise in Curves and Surfaces, second edition, by Montiel and Ros, which asks the reader to prove that the only surfaces of revolution with zero mean curvature are catenoids and planes.
After computing the components of the second fundamental form, for the parametrization $$ X(u, v) = (x(u), y(u) \cos v, -y(u) \sin v) $$ we arrive at the condition that the mean curvature is zero if and only if $$ y\left(\frac{dx}{du} \frac{d^2y}{du^2} - \frac{dy}{du}\frac{d^2x}{du^2} \right) - \frac{dx}{du} \left( \left(\frac{dx}{du} \right)^2 + \left(\frac{dy}{du} \right)^2 \right) = 0. \qquad \qquad (*) $$ Working on the set where $dx/du \neq 0$, the book argues that we can express $y$ as a function of $x$, since $x$ is invertible and hence, dividing by $(dx/du)^3$ we arrive at $$ y y'' - 1 - (y')^2 = 0, $$ where $y' = dy/dx$.
My question is:
What happened to the term $\frac{dy}{du}\frac{d^2x}{du^2}$ inside the first parenthesis in $(*)$?
Thanks in advance.
Both terms in the parentheses in ($\ast$) should contribute to $y''$, because when writing down $y''$ using the chain rule and product rule you'll see that it has two terms because $y'$ has two terms multiplied
$$y'=\frac{dy}{du}\frac{du}{dx}$$
You need to use the product rule. The form you'll get after taking another $x$ derivative of $y'$ however, is not the form that is in ($\ast$), you'll have to use the inverse function theorem to get it in the form in ($\ast$). Hope this helps.