Put $N\colon=\R\sm\{0\}\times\R\times(0,2\pi)$, $M$ three dimensional Minkowski space $\R_1^3$ and $F:N\rar M$:
$$
F(r,\theta,\vp)\colon=(r\cosh\theta,r\sinh\theta\cos\vp,r\sinh\theta\sin\vp)~.
$$
Compute the pull-back metric $h$ of $F:N\rar M$.
This time let us compute the metric coefficients directly: $y\colon=(r,\theta,\vp)$, $x\colon=F(y)$: The Jacobian of $F$ at $y$ is given by
$$
\left(\begin{array}{ccc}
\cosh\theta&r\sinh\theta&0\\
\sinh\theta\cos\vp&r\cosh\theta\cos\vp&-r\sinh\theta\sin\vp\\
\sinh\theta\sin\vp&r\cosh\theta\sin\vp&r\sinh\theta\cos\vp
\end{array}\right)
$$
and therefore
\begin{eqnarray*}
T_yF(E^r)&=&\cosh\theta\,E^0+\sinh\theta\cos\vp\,E^1+\sinh\theta\sin\vp\,E^2\\
T_yF(E^\theta)&=&r\sinh\theta\,E^0+r\cosh\theta\cos\vp\,E^1+r\cosh\theta\sin\vp\,E^2\\
T_yF(E^\vp)&=&-r\sinh\theta\sin\vp\,E^1+r\sinh\theta\cos\vp\,E^2
\end{eqnarray*}
This implies
\begin{eqnarray*}
h(E^r,E^r)&=&-\cosh^2\theta+\sin^2\vp\sinh^2\theta+\cos^2\vp\sinh^2\theta=-1\\
h(E^\theta,E^\theta)&=&-r^2\sinh^2\theta+r^2\cosh^2\theta\cos^2\vp+r^2\cosh^2\sin^2\vp=r^2\\
h(E^\vp,E^\vp)&=&r^2\sinh^2\theta\sin^2\vp+r^2\sinh^2\theta\cos^2\vp=r^2\sinh^2\theta
\end{eqnarray*}
and all off diagonal terms vanish.