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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 $DF=DbDa$, the product of a two-by-one and a one-by-two matrix. Its rank is at most one, so its Jacobian determinant is identically zero.

Conversely, continuity and $f_y(0,0)\ne 0$ give a neighborhood $U$ on which $f_y$ stays nonzero. The identity $f_x{g}_y-f_y{g}_x=0$ then implies $g_x=f_x{g}_y/f_y$. Therefore the smooth function $e=g_y/f_y$ satisfiesFor a graph curve $c(t)=(t,\alpha (t))$ with $F\circ c$ constant, the first component must have derivative zero. Thus the necessary equation isIt is also sufficient: it makes $d(f\circ c)/dt=0$, and then $d(g\circ c)/dt=ed(f\circ c)/dt=0$.

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.