= Solution
The <induced metric> is obtained from the ambient <Euclidean metric> by the <pullback of a Riemannian metric> operation. Differentiating $x^2+a^2\rho^2=1$ on a branch with $x\ne0$ gives $dx=-a^2\rho\,d\rho/x$. Also $dy^2+dz^2=d\rho^2+\rho^2d\phi^2$. Consequently
$$
\boxed{g=E(\rho)d\rho^2+\rho^2d\phi^2,\qquad E(\rho)=1+\frac{a^4\rho^2}{1-a^2\rho^2}=\frac{1+a^2(a^2-1)\rho^2}{1-a^2\rho^2}.}
$$
The assumption that $(\rho,\phi)$ are coordinates already excludes the equator $x=0$ as well as the poles $\rho=0$. On such a chart, $0<\rho<1/|a|$ and $E>0$.
The <conformal flattening of a surface of revolution> supplies the particularly simple <conformal factor>
$$
\boxed{\Omega=\rho^{-1},\qquad \Omega^2g=\frac{E(\rho)}{\rho^2}d\rho^2+d\phi^2=d\sigma^2+d\phi^2,\quad \sigma(\rho)=\int_{\rho_*}^{\rho}\frac{\sqrt{E(r)}}r\,dr.}
$$
Since $\sigma'>0$, these are <isothermal coordinates> with a flat rescaled metric. An angular coordinate is understood on a local branch; the flat metric may equally be viewed locally as a cylinder metric. The apparent divergence of $E$ at the equator is a coordinate failure, not a singularity of the <induced metric>. For example, $(x,\phi)$ give $g=[1+x^2/(a^2(1-x^2))]dx^2+(1-x^2)d\phi^2/a^2$, regular at $x=0$.
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