Proving that a Riemannian product is conformally flat We say that two metrics $g_1$ and $g_0$ on a Riemannian manifold $M^{n}$ are conformal, if there exists a positive function $\varphi\in C^{\infty}(M)$ such that \begin{equation}
g_1=\varphi^{2}g_0.
\end{equation}
In particular, we say that a Riemannian manifold $(M^{n},g_1)$ is conformally flat on $x\in M^{n}$, if there exists an open subset $U_{x}$ on $M^{n}$ that is conformal to an open subset of $\mathbb{R}^{n}$ endowed with an Euclidian metric $g_0$. We say that $(M,g_1)$ is conformally flat, if it is conformally flat on every point $x$.
My question is:
I would like to show that $\mathbb{H}^{2}\times \mathbb{R}=\{(x_1,x_2,x_3) \in \mathbb{R}^{3}; \, x_2>0\}$ is conformally flat  by finding a metric conformal to the Euclidean metric $g_0$ in $\mathbb{R}^{3}$.
Thank you for your support.
 A: Let's denote by $\mathbb{E}^1$ the Riemannian manifold which is the space $\mathbb{R}$ with its standard Euclidean metric. And denote by $\mathbb{H}^2$ the upper half plane model of the hyperbolic plane, which comes equipped with the Riemannian metric $\frac{1}{y^2}(dx^2 + dy^2)$. As stated in the comments, the metric on $\mathbb{H}^2 \times \mathbb{E}^1$ is
defined to be $\frac{1}{y^2}(dx^2 + dy^2) + dz^2$.
As $y^2$ is a positive smooth function on $\mathbb{H}^2 \times \mathbb{E}^1$,
$\frac{1}{y^2}(dx^2 + dy^2) + dz^2$ is conformally equivalent to
$dx^2 + dy^2 + y^2 dz^2$. Notice that $dy^2 + y^2 dz^2$ has the same form as the Euclidean plane metric in polar coordinates, $dr^2 + r^2 d\theta^2$.
So for each $z_0 \in \mathbb{R}$, the map $\phi(x, y, z) = (x, y', z') := (x, y \cos{z}, y \sin{z})$ determines an isometry from the open set $\{ y > 0, z_0 < z < z_0 + 2\pi \}$ in $(\mathbb{R}_{y > 0}^3, dx^2 + dy^2 + y^2 dz^2)$ to an open subset of $(\mathbb{R}^3, dx^2 + dy'^2 + dz'^2)$. So $\mathbb{H}^2 \times \mathbb{E}^1$ is conformally flat.
See pages 76-78 from the book on conformal geometry by Kulkarni and Pinkall for a characterization of conformal flatness of a product. There it is stated (and proved) that

A Riemannian product $(M_1, g_1) \times (M_2, g_2)$ is conformally
flat if and only if either
i) $(M_i, g_i)$ is $1$-dimensional, and $(M_j, g_j)$ ($j \neq i$) is a
Riemannian space-form,
or
ii) $(M_1, g_1)$ and $(M_2, g_2)$ are Riemannian space-forms of
dimension at least $2$, with non-zero opposite curvatures.

