\(O\) is the origin, \(\overrightarrow{OA} = \mathbf{a}\), and \(\overrightarrow{OB} = \mathbf{b}\). The point \(X\) is on \(AB\) such that \(AX = 2XB\) and \(Y\) is the midpoint of \(OX\).

Find \(\overrightarrow{OX}\) and hence show that \(\overrightarrow{BY} = \dfrac{1}{6}\mathbf{a}-\dfrac{2}{3}\mathbf{b}\).

\(BY\) produced meets \(OA\) at \(Z\). Using the facts that \[\overrightarrow{OZ} = \mathbf{b} + k\,\overrightarrow{BY} \quad \text{for some value of } k\] and \[\overrightarrow{OZ} = h\,\mathbf{a} \quad \text{for some value of } h,\] find the position vector of \(Z\).