How can I calculate the curl of a 2D field like $\textbf{F}= F_x(x,y)\textbf{i} + F_y(x,y)\textbf{j}$ if the curl is defined is 3D? My book says to apply the definition of curl to the associated 3D field $\textbf{F} = F_x(x,y)\textbf{i} + F_2(x,y)\textbf{j} +0\textbf{k}$, but I don't get the expected result which is $\text{rot} \textbf{F}= (\partial_x F_y - \partial_y F_x)\textbf{k}$
$\left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right)\mathbf {i} +\left({\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\right)\mathbf {j} +\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right)\mathbf {k} ={\begin{bmatrix}{\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\\{\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\\{\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\end{bmatrix}}$
since $F_z=0$ the result should be
$\text{rot} \mathbf{F} =\left(-{\frac {\partial F_{y}}{\partial z}}\right)\mathbf {i} +\left({\frac {\partial F_{x}}{\partial z}}\right)\mathbf {j} +\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right)\mathbf {k} $
The formula you state gives you the correct derivation of $\text{rot}\textbf{F}$, considering that $\frac{\partial \textbf{F}}{\partial z}=\vec{0}$ and $\nabla F_z=\nabla 0=\vec{0}$.
In your last equation, you have to consider that $\frac{\partial F_x}{\partial z}=\frac{\partial F_y}{\partial z}=0$, since $\textbf{F}$ does not depend on $z$