= Solution
The needed partial <derivatives> are
$$
\begin{aligned}
f_x&=\alpha\beta(1+\gamma_1\dot x+\gamma_2\dot y),&f_y&=0,\\
f_{\dot x}&=\alpha\gamma_1(1+\beta x)+\delta_1\dot x,&f_{\dot y}&=\alpha\gamma_2(1+\beta x)+\delta_2\dot y.
\end{aligned}
$$
Substitution into the <Euler-Lagrange equations> gives
$$
\alpha\beta\gamma_1\dot x+\delta_1\ddot x=\alpha\beta(1+\gamma_1\dot x+\gamma_2\dot y),\qquad\alpha\beta\gamma_2\dot x+\delta_2\ddot y=0.
$$
The $\gamma_1$ terms cancel in the first equation, leaving
$$
\boxed{\delta_1\ddot x=\alpha\beta(1+\gamma_2\dot y),\qquad\delta_2\ddot y=-\alpha\beta\gamma_2\dot x.}
$$
These are accompanied by $x(0)=y(0)=0$ and $x(T)=y(T)=S$. The cancellation is consistent with part (b): the term proportional to $\gamma_1$ is the <total derivative> of $\alpha\gamma_1(x+\beta x^2/2)$.
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