= Solution
If $F=b\circ a$ with $a:\mathbb R^2\to\mathbb R$ and $b:\mathbb R\to\mathbb R^2$, the <chain rule> gives $DF=Db\,Da$, the product of a two-by-one and a one-by-two <matrix>. Its <rank> is at most one, so \b[its <Jacobian determinant> is identically zero].
Conversely, <continuity> and $f_y(0,0)\ne0$ give a neighborhood $U$ on which $f_y$ stays nonzero. The identity $f_xg_y-f_yg_x=0$ then implies $g_x=f_xg_y/f_y$. Therefore the smooth function $e=g_y/f_y$ satisfies
$$
\boxed{\nabla g=e\nabla f\quad\hbox{on }U.}
$$
For a graph curve $c(t)=(t,\alpha(t))$ with $F\circ c$ constant, the first component must have derivative zero. Thus the necessary equation is
$$
\boxed{\alpha'(t)=-\frac{f_x(t,\alpha(t))}{f_y(t,\alpha(t))}.}
$$
It is also sufficient: it makes $d(f\circ c)/dt=0$, and then $d(g\circ c)/dt=e\,d(f\circ c)/dt=0$.
To justify the existence through every nearby point, choose a closed rectangle about the origin inside $U$. The right-hand side $h(t,y)=-f_x/f_y$ is smooth there, hence bounded and locally Lipschitz in $y$. The <Picard-Lindelöf theorem> supplies a unique local solution through every initial point $(t_0,y_0)$ in a smaller open rectangle $V$. The boundedness of $h$ and a positive distance from the smaller rectangle to the outer boundary permit a common sufficiently short time interval about each $t_0$ on which the solution stays in $U$. Each point of $V$ therefore lies on one of these graph curves, and the preceding chain-rule calculation proves that $F$ is constant along it. This proves <rank-one level curves from a characteristic differential equation> using local ODE existence. The curves are local in their time parameter; no unjustified global existence for all real $t$ is required.
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