Extracting an unknown PDE from a known charactersitc curve My question regards what I assume would be called the general method of characteristics for third-order equations.  I am looking for PDE for a function of two variables $x$ and $y$ having a characteristic curve in the form
$$
y^2=x^3+\alpha x+\beta~~.
$$
This type of curve is called an elliptic curve.  $\alpha$ and $\beta$ are two constants and the elliptic discriminant is such that the curve is non-singular (obviously).  I will make an example using the method of characteristics for first-order equations, and then I will say more precisely what I'm looking for once I have laid out the context.  Given a function $u(x,y)$ and
$$
\partial_x u+\alpha\,\partial_y u=0~~,
$$
Following what I understand to be the normal algorithmic procedure for solving such PDEs, I will introduce
$$
\dfrac{dy}{ds}=\alpha\qquad\qquad \dfrac{dx}{ds}=1\qquad\qquad \dfrac{dz}{ds}=0.
$$
Integration yields
$$
y(s)=as+c_1\qquad\qquad    x(s)=s+c_2       \qquad\qquad  z(s)=c_3~~.
$$
Eliminating $s$, we find the characteristic curves for the PDE are given by the following lines in $\mathbb{R}^3$:
$$
y-\alpha x=y_0\qquad\text{and}\qquad z=k~~.
$$
Here, $y_0$ and $k$ are two constants.  It follows that $z(x,y)$ is constant when $y-\alpha x$ is constant.  Setting $u=z$, we find $u(x,y)=f(y-\alpha x)$ is a solution to the stated PDE.
Now, my question: I want to find a PDE for $u(x,y)$ whose characteristic curve in the plane of those variables (meaning the analogue of $y-\alpha x=y_0$) is an elliptic curve.

$$y^2=x^3+\alpha x+\beta \tag 1$$
$$2y\frac{dy}{dx}=3x^2+\alpha$$
$$y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2=3x\tag 2$$
Eliminating $\alpha$ and $\beta$ from Eq.$(1)$ leadsto the ODE $(2)$.
This can be written equivalently on a particular form of PDE in which a dummy variable (for example $t$) appears but is not involved in the PDE :
$$u(x,t)=y(x)\quad\implies\quad u_x=\frac{dy}{dx}\quad\text{and}\quad u_{xx}=\frac{d^2y}{dx^2}$$
$$\boxed{y\:u_{xx}+(u_x)^2=3x}\tag 3$$
Equation $(1)$ is a particular solution among the infinity of solutions of $(3)$.
So, the answer to the original question is not unique but cannot be a first order linear PDE.
If we consider the first order PDE :
$$u_x+A(x,y)u_y=0$$
a characteristic equation is :
$$y-\int A(x,y)dx=\text{constant}$$
and the general solution is :
$$u(x,y)=f\left(y-\int A(x,y)dx\right)\quad\text{with arbitrary function }f.$$
or $u(x,y)=f(y-Ax)$ if $A$ is constant, as you correctly found it.
For Eq.$(1)$ to be a characteristic equation it is necessary that
$$\frac{dy}{dx}-A(x,y)=0 \quad\implies\quad 2y\frac{dy}{dx}=3x^2+\alpha=2yA(x,y)$$
Thus the first order linear PDE is :
$$u_x+\frac{1}{2y}(3x^2+2\alpha)u_y=0$$
$$2yu_x+(3x^2+2\alpha) u_y=0$$
Since the constant $\alpha$ remains, another differentiation is necessary to make it disappear, leading to a nonlinear second order PDE. Depending on the variable chosen to differentiate, different PDE can be obtained. This is not surprising since an infinity of second order PDEs are solution of the question.
