Obstructions for a metric to be conformally equivalent to a product metric Is there  a Riemannian metric on $\mathbb{R}^{2}=\mathbb{R} \times \mathbb{R}$ which is not conformally equivalent to a product metric?
More generally, assume that $M$  and $N$ are two manifolds. What obstructions are there for a metric $g$ on $M \times N$ to be conformally equivalent to a product metric for metrics 
$g_{1}$ and $g_{2}$ on $M$ and $N$, respectively?
 A: Every metric on $R^2$ is conformally equivalent to either the standard metric on $R^2$ or to the standard metric on the unit disc. Thus there are only two conformal equivalence classes. As every class contains a metric which is a product, every metric on $R^2$ is conformally equivalent to a product metric. Of course, this argument uses dimension $2$
in the very essential way.
EDIT. The answer for the $2D$ torus is also yes. Every metric on a torus is conformally
equivalent to a flat metric. About higher dimensions I do not know.
The general formulation of the Uniformization theorem is that every open simply connected Riemann surface is conformally equivalent to the disc or to the plane. Equivalent formulation is that every Riemannian metric on such Riemann surface is conformally equivalent to the flat metric on the plane or to the Poincare metric in the unit disc.
A corollary is that every Riemann surface has a conformal metric of constant curvature.
EDIT. To construct an example of a product metric in the plane which is conformally equivalent to the disc, use the criterion that every complete metric of curvature $<-k$,
with $k>0$
will be conformally equivalent to the unit disc. (This is called the Ahlfors-Schwarz Lemma). An explicit example of such product metric is $ds^2=e^{x^2}dx^2+e^{y^2}dy^2$. 
Just compute the curvature. 
The references are: Ahlfors, Conformal invariants; Hubbard, Teichmuller theory and applications to geometry, topology and dynamics (recommended!), and many other books on Riemann surfaces.
Let me add that in your case you know in advance that the surface is homeomorphic to the plane (or to the torus) and with this additional information the proof of the Univormization theorem can be substantially simplified. Perhaps the simplest proof (if you know in advance what your surface is homeomorphic to) is in Goluzin, Geometric function theory.  
