Given that

\(\displaystyle y=\frac{(3x-4)^{\frac13}}{2x+1},\)

show that \(\frac{\mathrm dy}{\mathrm dx}\) can be written in the form

\(\displaystyle \frac{Ax+B}{(2x+1)^2(3x-4)^{\frac23}},\)

where \(A\) and \(B\) are integers. Hence find the coordinates of the stationary point on the curve.