Conformal flatness of ellipsoid

[ergospherical]

Consider the ellipsoid:$$\mathcal{Q} := \{ \mathbf{x} \in \mathbb{R}^3, \ x^2 + a^2(y^2 + z^2) = 1 \}$$We have local coordinates $\chi^A = (\rho, \phi)$ on the ellipsoid surface defined by $y = \rho \cos{\phi}$ and $z = \rho \sin{\phi}$. First we look for the metric $\gamma := \phi^{*} g$ induced from the Euclidean metric, specifically:$$\gamma_{AB} = \frac{\partial x^{i}}{\partial \chi^A} \frac{\partial x^j}{\partial \chi^B} \delta_{ij}$$Using also that $x(\rho, \phi) = \sqrt{1-a^2 \rho^2}$, I obtain that the pull back of the metric is:$$\gamma = \left( 1 + \frac{a^4 \rho^2}{1-a^2 \rho^2} \right) d\rho^2 + \rho^2 d\phi^2$$We want to find a non-zero function $\Omega$ such that the conformal metric $\Omega^2 \gamma_{ij}$ is flat, i.e. that there is some transformation that brings $\Omega^2 \gamma_{ij}$ into a form resembling $\delta_{ij}$. I've had no luck with my guesses, but there must be an intuitive way of seeing this?