I was doing a question which got reduced to
Say we have$$ \frac{\partial^2 u}{\partial x \partial y}=F(u) $$ for some $F(u)$ . Show that the transformation $$ (x, y) \longrightarrow\left(c x, c^{-1} y\right), \quad c \in \mathbb{R} \backslash\{0\} $$ forms a symmetry group of the PDE and find the vector field generating this group.
Showing that it is a group of symmetries was easy enough. However, I am not sure what the vector field generating this group is. I have the following definitions
Defn (Generating Vector field with respect to a flow) We say that a vector field $\boldsymbol{V}: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is the generating vector field with respect to a flow $g^{\varepsilon}$ if $$ \boldsymbol{V}(\boldsymbol{x})=\left.\frac{d}{d \varepsilon} g^{\varepsilon} \boldsymbol{x}\right|_{\varepsilon=0} . $$
The motivation for this definition is that this is the leading order term in the Taylor expansion of the flow map: $$ g^{\varepsilon} \boldsymbol{x}=\boldsymbol{x}+\varepsilon \boldsymbol{V}(\boldsymbol{x})+o(\varepsilon) . $$
This flow map is an example of a
Defn (One-parameter group of transformations) A smooth $\operatorname{map} g^{\varepsilon}: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is called a one-parameter group of transformations (1.p.g.t) if $$ g^0=\mathrm{id}, \quad g^{\varepsilon_1} g^{\varepsilon_2}=g^{\varepsilon_1+\varepsilon_2} . $$
What we have here is not a one parameter group of transformations because we do not have either of the equalities.
Regardless, I tried to press on with the definition of a generating vector field which got me
$$ V(x,y)=(x,-y). $$
This doesn't seem to be quite right as this also corresponds to the lie point symmetry $$ g^{\varepsilon}:(x, y, u) \mapsto\left(e^{\varepsilon} x, e^{-\varepsilon} y, u\right). $$
Question: What is the definition of a vector field generating a group of symmetries in the context of this question?
There is nothing wrong in what you wrote. The generating vector field is $$ V(x, y)=(x, -y), $$ and the 1-parameter group is $$ g^t(x, y, u)=(e^t x, e^{-t}y, u). $$
If you are interested in directly verifying this, all you need to do is to solve the system of ODEs $$\begin{cases} x'(t)=V_1((x(t), y(t)))=x(t),\\ y'(t)=V_2((x(t), y(t)))=-y(t),\\ x(0)=x,\ y(0)=y, \end{cases} $$ where $V_1$ and $V_2$ are the components of $V$.
Here is a field plot of the vector field.
The flow lines are branches of the hyperbolas $xy=a$ for various constants $a$. This matches with your transformation $$ (x, y)\to (cx, c^{-1}y),\quad c\in {\mathbb R}\backslash \{0\}. $$