I am tackling this problem, from a 2016 Cambridge Astrophysics Tripos past paper:
Let $\Phi(R, z)$ be the axi-symmetric Galactic potential. At the Solar location, $(R, z) = (R_0, 0)$, prove that $$\frac{\partial^{2} \Phi}{\partial z^{2}}=4 \pi G \rho_{0}+2\left(A^{2}-B^{2}\right)$$ where $G$ is the gravitational constant, $\rho_0$ is the density in the Solar neighborhood and $A$ and $B$ are Oort’s constants.
What I do is:
Rewrite equation we want to prove
Note:
$A^2-B^2=-\frac{v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}$
Using definiton of $A$ and $B$. (Which are: $\left.A \equiv \frac{1}{2}\left(\frac{v_{c}}{R}-\frac{d v_{c}}{d R}\right)\right|_{R_{0}}$ and $B \equiv-\left.\frac{1}{2}\left(\frac{v_{c}}{R}+\frac{d v_{c}}{d R}\right)\right|_{R_{0}}$).
Rewrite what we want to prove:
$$ \bbox[5px,border:3px solid green]{ \frac{\partial^{2} \Phi}{\partial z^{2}}=4 \pi G \rho_{0}-2\frac{v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R} } $$
Use Poisson's equation:
$$\nabla^2\Phi=4\pi G\rho_0$$
Expand LHS of Poisson:
$$\frac{\partial^2}{\partial R^2}\Phi+\frac{\partial^2}{\partial z^2}\Phi=4\pi G \rho_0$$
Use equation of motion for circular orbit:
$$\frac{v_c^2}{R}=\frac{\partial}{\partial R}\Phi$$
Differentiate once more wrt $R$:
$$\frac{\partial^2}{\partial^2 R}\Phi=\frac{\partial}{\partial R}\frac{v_c^2}{R}=\frac{2v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}-\frac{1}{R^2}v_c^2$$
Collect results
Substitute result into Poisson's equation expanded form:
$$\frac{2v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}-\frac{1}{R^2}v_c^2+\frac{\partial^2}{\partial z^2}\Phi=4\pi G\rho_0$$
Rearrange:
$$ \bbox[5px,border:3px solid green]{ \frac{\partial^2}{\partial z^2}\Phi=4\pi G\rho_0-2\frac{v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}+\bbox[5px,border:3px solid red]{\frac{1}{R^2}v_c^2} }$$
Conclude
Compare with what we wanted to prove (ie the two green-boxed equations). We have an extra term (red boxed) which shouldn't be there. How do I fix the proof?
Key: be careful what the Laplacian is in different coordinate systems.
Poission equaiton: $$\nabla^{2} \Phi=4 \pi G S_{0}$$
In cylindrical polar coordinates: $$\frac{1}{r} \frac{d}{dr}\left(r \frac{d \Phi}{d r}\right)+\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho_{0}$$
We also know from $\ddot{\vec{r}}=-\nabla\Phi$ that $\ddot{r}-r \dot{\phi}^{2}=-\frac{\partial\Phi}{\partial r}$, so for circular orbit, we have: $r\dot{\phi}^2=\frac{\partial \Phi}{\partial r}$. This means that we can write $v_c=r^2\dot{\phi}^2=r\frac{\partial \Phi}{\partial r}$
Rewrite Poisson above:
$$\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho_{0}-\frac{1}{r} \frac{d}{d r}\left(v_{c}^{2}\right) = 4 \pi G \rho_{0}- \frac{2v_c}{r}\frac{dv_c}{dr}$$
Keeping in mind that $A^2-B^2=-\frac{v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}$, rewrite RHS:
$$\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho_{0}+2(A^2-B^2)$$
which is what we wanted.