Show that \begin{align*} ( (-1, 0, 0)\times F(x_0, y_m, z_m)\Delta y\Delta z +(1, 0, 0)\times F(x_0+\Delta x, y_m, z_m)\Delta y\Delta z+(0, -1, 0)\times F(x_m, y_0, z_m)\Delta x\Delta z +(0, 1, 0)\times F(x_m, y_0+\Delta y, z_m)\Delta x\Delta z+(0, 0, -1)\times F(x_m, y_m, z_0)\Delta x\Delta y+(0, 0, 1)\times F(x_m, y_m, z_0+\Delta z)\Delta x\Delta y) \frac{1}{\Delta x \Delta y\Delta z} \end{align*} They tend to $curl\: F(x_0, y_o, z_0)$ when $\Delta x, $ $\Delta y$, $\Delta z$ tend to zero.
The other calculations are missing, but it has become complicated. Since \begin{align*} (-1, 0, 0)\times F(x_0, y_m, z_m)\Delta y\Delta z +(1, 0, 0)\times F(x_0+\Delta x, y_m, z_m)\Delta y\Delta z = \\ \left( (0,f_3(x_0,y_m,z_m),-f_2(x_0,y_m,z_m)\right)\Delta y\Delta z +\left( 0,-f_3(x_0+\Delta x,y_m,z_m),f_2(x_0+\Delta x,y_m,z_m) \right)\Delta y\Delta z \\ =\left( 0,f_3(x_0,y_m,z_m)-f_3(x_0+\Delta x,y_m,z_m),f_2(x_0+\Delta x,y_m,z_m)-f_2(x_0,y_m,z_m) \right)\Delta y \Delta x \end{align*} Applying the mean value theorem, there exist $\bar{x}$ y $\bar{\bar{x}}$, with $x_0<\bar{x},\bar{\bar{x}}<x_0+\Delta x$, such that the above expression is equal to \begin{align*} \left(0,-\Delta x\frac{\partial}{\partial x}f_3(\bar{x},y_m,z_m),\Delta x\frac{\partial}{\partial x}f_3(\bar{\bar{x}},y_m,z_m) \right)\Delta y\Delta z = \left(0,-\frac{\partial}{\partial x}f_3(\bar{x},y_m,z_m),\frac{\partial}{\partial x}f_3(\bar{\bar{x}},y_m,z_m) \right)\Delta x\Delta y\Delta z \end{align*}
Similarly \begin{align*} \left(-\frac{\partial}{\partial y}f_1(x_m,\bar{y},z_m),0,\frac{\partial}{\partial y}f_1(x_m,\bar{\bar{y}},z_m) \right)\Delta x\Delta y\Delta z \\ \left(-\frac{\partial}{\partial z}f_2(x_m,y_m,\bar{z}),\frac{\partial}{\partial z}f_2(x_m,y_m,\bar{\bar{z}}),0 \right))\Delta x\Delta y\Delta z \\ \end{align*} Well, I don't know if this is right