This is a part of the derivation of a Runge-Kutta method. Consider the equation $$y' = f(t,y)$$
then $$y'' = f_t+fy\cdot y'$$
$$y'''=f_{tt}+f_{ty}f+(f_t+f_yf)f_y+f(f_{ty}+f_{tt}f)$$
where $f_t,f_y$ denote partial derivatives. Instead of talking about derivatives, we are considering differentials. Fine, but how do we arrive at this expression for $y'''$? Supposedly the chain rule is at work here, but I don't see it.
Edit : This is from the textbook, "Numerical analysis, Kincaid and Cheney". Specifically page 499.
It looks wrong... First insert the ODE in the second derivative $$ y''=f_t+f_yf. $$ Then apply the same total differentiation by $t$ \begin{align} y'''&=(f_t+f_yf)_t+(f_t+f_yf)_yy' \\ &=(f_{tt}+f_{yt}f+f_yf_t)+(f_{ty}+f_{yy}f+f_yf_y)y' \\ &=f_{tt}+f_{yt}f+f_yf_t+(f_{ty}+f_{yy}f+f_yf_y)f \end{align} I do not see any $f_t^2$ or $ff_{tt}f$ terms occurring. Collecting equal terms gives $$ y'''=f_{tt}+2f_{yt}f+f_yf_t+f_{yy}f^2+f_yf_yf $$ Note that the next derivative needs a more careful treatment if $y$ is not scalar, as terms like $f_yf_{yy}[f,f]$ and $f_{yy}[f_yf,f]$ are no longer the same.