= Solution
{c}
The required condition is $D_2F(a,b)\neq0$. The <implicit function theorem> then gives a unique local differentiable branch $y=f(x)$ with $f(a)=b$ and
$$
\boxed{f'(x)=-\frac{D_1F(x,f(x))}{D_2F(x,f(x))}.}
$$
The denominator remains nonzero after shrinking the neighborhood by <continuity>.
For the specified smooth function,
$$
F_x=y+\cos(x+y),\qquad F_y=x+\cos(x+y).
$$
At $(0,0)$ both equal one, so the condition holds and $f'(0)=-1$. Differentiate $F(x,f(x))=0$ twice:
$$
F_{xx}+2F_{xy}f'+F_{yy}(f')^2+F_yf''=0.
$$
At zero, $F_{xx}=F_{yy}=0$, $F_{xy}=1$, and $F_y=1$. Consequently
$$
\boxed{f'(0)=-1,\qquad f''(0)=2.}
$$
The additional derivatives exist because this particular $F$ is smooth; the general first-derivative theorem only requires the stated continuous first <partial derivatives>.
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