Deriving $f\frac{\partial u}{\partial z}$:

Take $\frac{\partial}{\partial z}$ of the $y$-momentum equation (Eq. ):

$$\frac{\partial}{\partial z}\left(fu\right) = \frac{\partial}{\partial z}\left(-\frac{1}{\rho_0}\frac{\partial p}{\partial y}\right)$$

Since $f$ is typically taken as the Coriolis parameter ( $2\Omega \sin \phi$), it only depends on latitude (and thus $y$ in a mid-latitude $\beta$-plane approximation) but is constant with $z$. Therefore:

$$f\frac{\partial u}{\partial z} = -\frac{1}{\rho_0}\frac{\partial}{\partial y}\left(\frac{\partial p}{\partial z}\right)$$

Now, substitute the hydrostatic relation (Eq. ) for $\frac{\partial p}{\partial z}$:

$$f\frac{\partial u}{\partial z} = -\frac{1}{\rho_0}\frac{\partial}{\partial y}\left(-\rho g\right)$$

Since $g$ is constant, we move it outside the derivative:

$$f\frac{\partial u}{\partial z} = \frac{g}{\rho_0}\frac{\partial \rho}{\partial y}$$

Finally, substitute $\rho = \rho_0 + \rho'$. Since $\rho_0$ is constant, $\frac{\partial \rho}{\partial y} = \frac{\partial \rho_0}{\partial y} + \frac{\partial \rho'}{\partial y} = \frac{\partial \rho'}{\partial y}$:

$$\mathbf{f\frac{\partial u}{\partial z} = \frac{g}{\rho_0} \frac{\partial \rho'}{\partial y}}$$