= Solution
{c}
Put $h(s)=\sin x(s)-\cos y(s)$. The <chain rule> gives
$$
h'=x'\cos x+y'\sin y,\qquad
f'=-h'\sin h=-(x'\cos x+y'\sin y)g.
$$
Apply the <product rule> and the <chain rule> once more: $g'=h'\cos h=h'f$, and
$$
h''=x''\cos x-(x')^2\sin x+y''\sin y+(y')^2\cos y.
$$
Consequently $f''=-(h')^2f-h''g$, so the requested coefficients are
$$
\boxed{\begin{aligned}
a&=-(x'\cos x+y'\sin y)^2,\\
b&=-x''\cos x+(x')^2\sin x-y''\sin y-(y')^2\cos y.
\end{aligned}}
$$
This is the <second derivative chain rule> along the <parametric curve> $(x(s),y(s))$.
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