Derivation of the Curl formula in cartesian coordinates. By calculating the circulation per area of a vector field 
$$F(x,y,z) = F_x(x,y,z)\vec{x} + F_y(x,y,z)\vec{y} + F_z(x,y,z)\vec{z}$$
in a small rectangle around $(x_0, y_0, z_0)$ on the $xy$ plane, it can be shown the limit as the sides of the rectangle approach zero is
$$\left(\frac{\partial F_y(x_0, y_0, z_0)}{\partial x} - \frac{\partial F_x(x_0, y_0, z_0)}{\partial y}\right)$$ 
The same calculation however is not that straightforward if the rectangle does not lie in the $xy$, $yz$, or $xz$ planes. Now if $\vec{n}$ is the normal of the plane, I thought that by performing a change of basis such that $\vec{n} \rightarrow \vec{z'} $ and by following the previous calculations we could show that the limit of the circulation per area is 
$$ \left(\frac{\partial F_{y'}(x'_0, y'_0, z'_0)}{\partial x'} - \frac{\partial F_{x'}(x_0, y_0, z_0)}{\partial y'}\right) $$
This is also the inner product of the curl of the vector field and the normal $\vec{n}$
As such the two should be equal:
$$\left(\frac{\partial F_{y'}(x'_0, y'_0, z'_0)}{\partial x'} - \frac{\partial F_{x'}(x'_0, y'_0, z'_0)}{\partial y'}\right) = \\
\left[\left(\frac{\partial F_z(x_0, y_0, z_0)}{\partial y} - \frac{\partial F_y(x_0, y_0, z_0)}{\partial z} \right)\vec{x} +
\left(\frac{\partial F_z(x_0, y_0, z_0)}{\partial x} - \frac{\partial F_x(x_0, y_0, z_0)}{\partial z} \right)\vec{y} + 
\left(\frac{\partial F_y(x_0, y_0, z_0)}{\partial x} - \frac{\partial F_x(x_0, y_0, z_0)}{\partial y} \right)\vec{z}\right] \cdot \vec{n} $$
I've been trying to prove the above equality for some time without success, specifically I am not sure how to handle the transformations correctly. Any help with this is much appreciated!
 A: A simpler approach is via integral theorems.
As stated in the question, the special cases for a rectangle in the $xy$ , $yz$ , $zx$ planes are well understood.
According to Green's theorem :
$$
\begin{cases}
\iint_{xy} \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) dx\, dy
= \oint_{xy} \left( F_x\, dx + F_y\, dy \right) \\
\iint_{yz} \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) dy\, dz
= \oint_{xy} \left( F_y\, dy + F_z\, dz \right) \\
\iint_{zx} \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) dz\, dx
= \oint_{xy} \left( F_z\, dz + F_x\, dx \right) \end{cases}
$$
But instead of rectangles, we take half rectangles, or better: the triangles $OAB$ , $OBC$ , $OAC$ respectively:

Thanks to Green's theorem we can replace area integrals by line-integrals; mind that they are counter-clockwise.
Then it is clear that, irrespective of any further content:
$$
\oint_{OAB} + \oint_{OBC} + \oint_{OAC} + \oint_{ABC} = 0
$$
Assuming that the operator rot(ation) is not defined yet in general, this means that we now have an expression for it:
$$
2 \iint_{ABC} \vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n}\, dA = \\
- \iint_{xy} \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) dx\, dy
- \iint_{yz} \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) dy\, dz
- \iint_{zx} \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) dz\, dx
$$
Continuing with infinitesimal volumes / areas and flipping normals on the right hand side, so that they become
the components of the normal at the left hand side:
$$
\vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n}\, \Delta A = \\
\left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right)\cdot n_x\, \Delta A +
\left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right)\cdot n_y\, \Delta A +
\left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\cdot n_z\, \Delta A 
$$
Leaving out the infinitesimal area $\,\Delta A\,$ gives us the same answer as found by the OP themselves.
A somewhat neater approach is to calculate mean values and let the area of the (red) triangle go to zero:
$$
\vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n} = \lim_{ABC \to 0}
\frac{\iint_{ABC} \vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n}\, dA}{\iint_{ABC} dA}
$$
Note. I've encountered essentially the same method at several places elsewhere in physics
(I think it's with stress and strain). Aanyway, a related subject is :
What does shear mean?
A: It is possible to prove this by taking Green's theorem and applying rotations. So what we do is rotate the whole space so that the normal is in the z direction (or whatever Green's Theorem configuration we like) and then use the rotation information to get an expression for the original space.
We start by applying a rotation around the x and y axis
$$ 
\left(\begin{array}{c} x' \\ y' \\ z' \end{array}\right) = 
   \left(\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i\end{array}\right) \cdot 
\left(\begin{array}{c} x \\ y \\ z\end{array}\right)
$$
This rotates the surface so that its normal at the required point, points upwards. This means,
$$\left(\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right) = \left(\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i\end{array}\right) \cdot \vec{n} 
$$ 
and by using the inversion property of rotation matrices,
$$
 \vec{n} =  \left(\begin{array}{ccc} a & d & g \\ b & e & h \\ c & f & i\end{array}\right) \cdot \left(\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right) = \left(\begin{array}{c} g \\ h \\ i \end{array}\right)
 $$
Notice that the normal $\vec{n}$ is the last row of our rotation matrix.
Since the first and second row are also unit vectors orthogonal to $\vec{n}$ we can write $\vec{n}$ as their cross product, 
$$ \vec{n} = \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ a & b & c \\ d & e & f \end{array}\right| 
$$
and hence, 
$$
n_x = \left|\begin{array}{cc}b & c \\ e & f \end{array}\right|
n_y = \left|\begin{array}{cc}d & f \\ a & c \end{array}\right|,
n_z = \left|\begin{array}{cc}a & b \\ d & e \end{array}\right|$$
We also want to rotate our vector field appropriately:
$$ 
\left(\begin{array}{c} F_{x'}(P') \\ F_{y'}(P') \\ F_{z'}(P') \end{array}\right) = 
   \left(\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i\end{array}\right) \cdot 
\left(\begin{array}{c} F_x(P) \\ F_y(P) \\ F_z(P)\end{array}\right)
$$
Finally we can derive our expression using simple determinant properties
and the linear relationships that we wrote above,
$$\begin{gather}\frac{\partial F_{y'}}{\partial x'} - \frac{\partial F_{x'}}{\partial y'} \end{gather} = \\
\left|\begin{array}{cc} \frac{\partial}{\partial x'} & \frac{\partial}{\partial y'} \\ F_{x'} & F_{y'} \end{array}\right| = \\
\left|\begin{array}{ccc} a\frac{\partial}{\partial x} + b\frac{\partial}{\partial y} + c\frac{\partial}{\partial z} & d\frac{\partial}{\partial x} + e\frac{\partial}{\partial y} + f\frac{\partial}{\partial z} \\ aF_{x} + bF_{y} + cF_{z} & dF_{x} + eF_{y} + fF_{z} \end{array}\right| = \\
\left| \begin{array}{cc} a\frac{\partial}{\partial x} & e\frac{\partial}{\partial y} \\ aF_x & eF_y\end{array}\right| + 
\left| \begin{array}{cc} a\frac{\partial}{\partial x} & f\frac{\partial}{\partial z} \\ aF_x & fF_z\end{array}\right| + 
\left| \begin{array}{cc} b\frac{\partial}{\partial y} & d\frac{\partial}{\partial x} \\ bF_y & dF_x\end{array}\right| + \\
\left| \begin{array}{cc} b\frac{\partial}{\partial y} & f\frac{\partial}{\partial z} \\ bF_y & fF_z\end{array}\right| + 
\left| \begin{array}{cc} c\frac{\partial}{\partial z} & d\frac{\partial}{\partial x} \\ cF_z & dF_x\end{array}\right| + 
\left| \begin{array}{cc} c\frac{\partial}{\partial z} & e\frac{\partial}{\partial y} \\ cF_z & eF_y\end{array}\right| = \\
(bf-ce)\left| \begin{array}{cc} \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_y & F_z \end{array}\right| + 
(af-cd)\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\ F_x & F_z \end{array}\right| + 
(ae-db)\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ F_x & F_y \end{array}\right| = \\
\left|\begin{array}{cc}b & c \\ e & f \end{array}\right|\left| \begin{array}{cc} \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_y & F_z \end{array}\right| -
\left|\begin{array}{cc}d & f \\ a & c \end{array}\right|\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\ F_x & F_z \end{array}\right| + 
\left|\begin{array}{cc}a & b \\ d & e \end{array}\right|\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ F_x & F_y \end{array}\right| = \\
n_x\left| \begin{array}{cc} \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_y & F_z \end{array}\right| - 
n_y\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\ F_x & F_z \end{array}\right| + 
n_z\left| \begin{array}{cc} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ F_x & F_y \end{array}\right| = \\
\left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{array}\right| \cdot \vec{n}$$
