A relevant reference is found here:
It is advised to absorb this one-dimensional theory first, before proceeding to 2-D.
$c)\; X = -y\, \partial_x+x\, \partial_y$
Disclaimer. In our (LaTeX) notes we have $f$ instead of $F$ ,
$(x_1,y_1)$ instead of $(\hat{x},\hat{y})$ , $\theta$ instead of $\varepsilon$ ,
$k$ instead of $j$ , and more. I didn't replace notations because my eyes
are bad and it is expected that the danger of making mistakes is greater than the advantage of being consistent with the question.
An example of a Continuous Transformation in two dimensions is a Rotation over an angle $\theta$:
$$
\left\{
\begin{array}{c}
x_1 = \cos(\theta) . x - \sin(\theta) . y \\
y_1 = \sin(\theta) . x + \cos(\theta) . y
\end{array}
\right.
$$
It might be asked how rotation of the coordinate system works out for a
function of these variables. With other words, how the following function would
be expanded as a Taylor series expansion around the original $f(x,y)$:
$$
f_1(x,y) = f(x_1,y_1)
= f(\,\cos(\theta).x - \sin(\theta).y\, , \,
\sin(\theta).x + \cos(\theta).y\, )
$$
Define other (polar) variables $(r,\phi)$ as:
$$
x = r.\cos(\phi) \quad \mbox{and} \quad y = r.\sin(\phi)
$$
Giving for the transformed variables:
$$
x_1 = r.\cos(\phi).\cos(\theta) - r.\sin(\phi).\sin(\theta) = r.\cos(\phi+\theta)
\\
y_1 = r.\cos(\phi).\sin(\theta) + r.\sin(\phi).\cos(\theta) = r.\sin(\phi+\theta)
$$
We see that $\phi$ is a proper canonical variable. Another function $g(\phi)$ is defined with this canonical variable as the independent one:
$$
g(\phi) = f(\,r.\cos(\phi)\, ,\,r.\sin(\phi)\,) = f(x,y)
$$
Now rotating $f(x,y)$ over an angle $\theta$ corresponds with a translation of
$g(\phi)$ over a distance $\theta$. Therefore $g(\phi+\theta)$ can be developed
into a Taylor series around the point of departure:
$$
g(\phi+\theta) = g(\phi) + \theta.\frac{dg(\phi)}{d\phi}
+ \frac{1}{2} \theta^2.\frac{d^2g}{d\phi^2} + ...
$$
Working back to the original variables $(x,y)$ with a well known chain rule for
partial derivatives:
$$
\frac{dg}{d\phi} = \frac{\partial g}{\partial x}\frac{dx}{d\phi}
+ \frac{\partial g}{\partial y}\frac{dy}{d\phi}
$$
Where:
$$
\frac{dx}{d\phi} = - r.\sin(\phi) = - y
\quad \mbox{and} \quad
\frac{dy}{d\phi} = + r.\cos(\phi) = + x
\quad \Longrightarrow
\\
\frac{dg}{d\phi} = \frac{\partial g}{\partial x}.(-y) + \frac{\partial g}{\partial y}.(+x)
\quad \Longrightarrow \quad
\frac{d}{d\phi} = x.\frac{\partial}{\partial y} - y.\frac{\partial}{\partial x}
$$
Herewith we find that $X = (x.\frac{\partial}{\partial y} - y.\frac{\partial}{\partial x})$ is the infinitesimal operator for Plane Rotations.
It is equal to differentiation with respect to the canonical variable, as expected. The end-result is:
$$
f_1(x,y) = \sum_{k=0}^{\infty} \frac{1}{k!}
\left[ \theta \left(x.\frac{\partial}{\partial y} - y.\frac{\partial}{\partial x}\right) \right]^k f(x,y)
= e^{ \theta (x\, \partial / \partial y - y\, \partial / \partial x) } f(x,y)
$$
This is true for any function $f(x,y)$. In particular, the independent
variables themselves can be conceived as such functions. Which means that:
$$
x_1 = e^{ \theta (x\, \partial / \partial y - y\, \partial / \partial x) } x \quad \mbox{and} \quad
y_1 = e^{ \theta (x\, \partial / \partial y - y\, \partial / \partial x) } y
$$
It is easily demonstrated that:
$$
(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}) x = - y
\quad \mbox{and} \quad (x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}) y = x
$$
Herewith we find:
$$
\sum_{k=0}^{\infty} \left[ \theta (x.\frac{\partial}{\partial y} - y.\frac{\partial}{\partial x}) \right]^k x = 1
- \theta.y - \frac{1}{2} \theta^2.x + \frac{1}{3!} \theta^3.y
+ \frac{1}{4!} \theta^4.x + ...
\\
= \cos(\theta).x - \sin(\theta).y = x_1
$$
Likewise we find:
$$
\sum_{k=0}^{\infty} \left[ \theta (x.\frac{\partial}{\partial y} - y.\frac{\partial}{\partial x} ) \right]^k y = 1
+ \theta.x - \frac{1}{2} \theta^2.y - \frac{1}{3!} \theta^3.x
+ \frac{1}{4!} \theta^4.y + ...
\\
= \sin(\theta).x + \cos(\theta).y = y_1
$$
Thus, indeed, the formulas for a far-form-infinitesimal rotation over an finite
angle $\theta$ can be reconstructed from the expansions.
$a)\; X = x\, \partial_x-y\, \partial_y$
Read the 1-D reference. We have the following results there:
$$
e^{ln(\lambda) \,x \frac{d}{dx}} f(x) = f(\lambda\,x)
$$
Where $\lambda$ is a positive
scaling factor. We also have:
$$
e^{-ln(\lambda) \,x \frac{d}{dx}} f(x) = e^{ln(1/\lambda) \,x \frac{d}{dx}} f(x) = f(x/\lambda)
$$
These results translate to 2-D in the following manner:
$$
e^{ln(\lambda) \,x \frac{\partial}{\partial x}} f(x,y) = f(\lambda\,x,y) \\
e^{-ln(\lambda) \,y \frac{\partial}{\partial y}} f(x,y) = f(x,y/\lambda)
$$
The two exponents are commutative, so we can write,
with $\;\ln(\lambda)=\mu\;\Longrightarrow\;\lambda=e^\mu=\exp(\mu)$ :
$$
e^{\mu(x\, \partial_x - y\, \partial_y)}\; f(x,y) = f\left(e^\mu x,e^{-\mu} y\right)
$$
In particular, with $\;X = x \frac{\partial}{\partial x} - y \frac{\partial}{\partial y}$ :
$$
\exp(\mu X) x = exp(\mu) x \quad \mbox{and} \quad \exp(\mu X) y = \exp(-\mu) y
$$
$b)\; X = x^2\,\partial_x+xy\,\partial_y$
As for this case, I don't see how we can say more than, in the OP's notation:
$$
F(\hat{x},\hat{y})=\sum_{j=0}^{\infty}\frac{\varepsilon^j}{j!}X^j F(x,y) = \exp{(\varepsilon X)} F(x,y)
$$
Then it follows that, for special functions $F(x,y)=x$
or $F(x,y)=y$ :
$$
\hat{x}=\exp{(\varepsilon X)}x \\ \hat{y}=\exp{(\varepsilon X)}y
$$
Please don't tell me that's all you want ..
