Your kinematic equation, which links TCP position to joint position is:
$$X = A_{0,3} \times Q$$
or
$$ \begin{pmatrix} x \\ y\\ z\\ 1\end{pmatrix} = A_{0,3} \times \begin{pmatrix} \theta_1 \\ \theta_2\\ \theta_3\\ 1\end{pmatrix} $$
based on this, you will need to write:
$$ \left\{
\begin{eqnarray}
x &=& f_x(\theta_1, \theta_2, \theta_3)\\
y &=& f_y(\theta_1, \theta_2, \theta_3)\\
z &=& f_z(\theta_1, \theta_2, \theta_3)\\
\alpha &=& f_\alpha(\theta_1, \theta_2, \theta_3)\\
\beta &=& f_\beta(\theta_1, \theta_2, \theta_3)\\
\gamma &=& f_\gamma(\theta_1, \theta_2, \theta_3)
\end{eqnarray}
\right.$$
where $\alpha, \beta, \gamma $ are the Euler angles of end-effector orientation (it does not matter in which order they are defined, you just have to pick one and stick with it, e.g. XYZ order or ZXZ order). Based on these equations you can write the Jacobian as:
then
$$ J = \left[
\begin{matrix}
\frac{\partial f_x}{\partial \theta_1} & \frac{\partial f_x}{\partial \theta_2} & \frac{\partial f_x}{\partial \theta_3}\\
\frac{\partial f_y}{\partial \theta_1} & \frac{\partial f_y}{\partial \theta_2} & \frac{\partial f_y}{\partial \theta_3}\\
\frac{\partial f_z}{ \partial \theta_1 } & \frac{\partial f_z}{\partial \theta_2} & \frac{\partial f_z}{\partial \theta_3}\\
\frac{\partial f_\alpha}{ \partial \theta_1 } & \frac{\partial f_z}{\partial \theta_2} & \frac{\partial f_z}{\partial \theta_3}\\
\frac{\partial f_\beta}{ \partial \theta_1 } & \frac{\partial f_z}{\partial \theta_2} & \frac{\partial f_z}{\partial \theta_3}\\
\frac{\partial f_\gamma}{ \partial \theta_1 } & \frac{\partial f_z}{\partial \theta_2} & \frac{\partial f_z}{\partial \theta_3}\\
\end{matrix}
\right]$$
Please note that this is $6x3$ as it is used to compute both linear and rotational velocities of the end-effector based on the 3 joint velocities as follows:
$$ \left[ \begin{array}
\dot{x} \\ \dot{y} \\ \dot{z} \\ \dot{\alpha} \\ \dot{\beta} \\ \dot{\gamma}
\end{array} \right]
= J \times \left[
\begin{array}
\dot{\theta_1} \\ \dot{\theta_2} \\ \dot{\theta_3}
\end{array}
\right]
$$