Ended up spending all day on this, but finally understand the solution! (Found this question based on a exercise problem in An Introduction to Nonlinear Partial Differential Equations by J. David Logan)
Anyways, here's the solution:
Really, the solution is not difficult, but can be difficult to see or easy to confuse if you're not used to it (as I was).
We know: $u_x=\xi$ and $u_y=\eta$. (ignoring brackets for ease of reading)
From that, we can easily see that
$$u_{xx}=\xi_x\\
u_{xy}=\xi_y=\eta_x\\
u_{yy}=\eta_y$$
This is where I had trouble continuing.
This trick is to invert the partials acting on the new coordinates in order to get them instead acting on the old coordinates (Ex. $\xi_x \rightarrow x_\xi=\phi_{\xi\xi}$).
But note that you cannot simply inverse the derivative as you could if there was only one dependent variable($ \xi_x \neq \frac{1}{x_\xi}$), But instead have to complete the matrix inversion (this is really explained well here). (Also, this question expresses changing the partial derivatives in a detailed form)
Therefore, the algebra becomes as follows:
$$
\begin{bmatrix}
u_{xx} & u_{xy} \\
u_{yx} & u_{yy} \\
\end{bmatrix}
=
\begin{bmatrix}
\xi_x & \xi_y \\
\eta_x & \eta_y \\
\end{bmatrix}
=
{
\begin{bmatrix}
x_\xi & x_\eta \\
y_\xi & y_\eta \\
\end{bmatrix}
}^{-1}
=
{
\begin{bmatrix}
\phi_{\xi\xi} & \phi_{\xi\eta} \\
\phi_{\eta\xi} & \phi_{\eta\eta} \\
\end{bmatrix}
}^{-1}
=
\frac{1}{\phi_{\eta\eta}\phi_{\xi\xi} -\phi_{\eta\xi}\phi_{\xi\eta}}
\begin{bmatrix}
\phi_{\eta\eta} & -\phi_{\eta\xi} \\
-\phi_{\xi\eta} & \phi_{\xi\xi} \\
\end{bmatrix}
$$
Assuming $u_{xy}=u_{yx}$ and $\phi_{\xi\eta}=\phi_{\eta\xi}$, and letting $\Delta=\frac{1}{\phi_{\eta\eta}\phi_{\xi\xi} -\phi_{\eta\xi}\phi_{\xi\eta}}$. We can see that:
$$
u_{xx} = \Delta\phi_{\eta\eta}\\
u_{xy} = -\Delta\phi_{\eta\xi}\\
u_{yy} = \Delta\phi_{\xi\xi}\\
$$
plugging this into the original equation:
$$
a(\xi,\eta)\Delta\phi_{\eta\eta} - 2b((\xi,\eta)\Delta\phi_{\eta\xi} + c((\xi,\eta)\Delta\phi_{\xi\xi} = 0
$$
and finally, dividing by $\Delta$, assuming $\Delta\neq 0$
$$
a(\xi,\eta)\phi_{\eta\eta} - 2b(\xi,\eta)\phi_{\eta\xi} + c(\xi,\eta)\phi_{\xi\xi} = 0
$$
For reference, another page I used as reference while trying to figure this out: 1
