For a PDE of the form:
$$a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u)$$
such that: $a,b,c\in C^1(\mathbb{R}^3),\ a\neq 0$, do exist $\phi(x,y,u),\psi(x,y,u)\in C^1(\mathbb{R}^3)$ that satisfy:
$$\nabla\phi\neq q\cdot\nabla\psi,\quad q\in\mathbb{R} \\
\nabla\phi\cdot(a,b,c)=\nabla\psi\cdot(a,b,c)=0? $$
The second requirement is the definition for integral surfaces, and the first requirement applies that if two such surfaces exist, then every solution of the PDE can be described using Lagrange's method, for some $F\in C^1$ that satisfies Lagrange's conditions.
Therefore, if I could find a PDE with continuous coefficients and $a\neq 0$ with a solution that can't be achieved using Lagrange's method, it will be a counterexample.
However, I couldn't find such an example, and I believe the claim is correct.
I found out that along characteristic lines: $\frac{d\psi}{dx}=\psi_x+\psi_y \frac{dy}{dx}+\psi_u \frac{du}{dx}=\psi_x+\frac{b}{a}\psi_y+\frac{c}{a} \psi_u=0$, same for $\frac{d\psi}{dx}=0$.
I need to find $\psi,\phi$ such that their gradients lie in the perpendicular plane of the characteristic line at every point. I'm not sure how to continue from here, though.