One knows that hyperbolic manifolds are locally conformally flat.
How about those negatively pinched manifolds, i.e. the sectional curvature $K$ satisfy:
$$
-\Lambda \le K \le -\lambda$$
for $\Lambda>\lambda$.


How about sufficiently pinched, i.e. $\Lambda/\lambda=1+\epsilon$ for $\epsilon$ small?


Are they have vanished Pontryagin classes?



Are negatively pinched manifold locally conformally flat?
No in dimensions $\geq 3$. To be conformally flat in 3 dimensions, the [Cotton tensor](http://en.wikipedia.org/wiki/Cotton_tensor) must to vanish, and in dimensions $\geq 4$, the [Weyl tensor](http://en.wikipedia.org/wiki/Weyl_tensor) must vanish. 


Maybe the point of your question though is to ask why is the Cotton or Weyl tensor non-vanishing? I don't have a good explanation for this. 


Here's a special example in 3 dimensions. Consider the metric $$dr^2+e^{2k\_1r}dx\_1^2+e^{2k\_2r}dx\_2^2.$$
 This metric is homogeneous, and when $k\_1>0, k\_2>0, k\_1\neq k\_2$, the metric is not conformally flat. The 3 principal sectional curvatures are $-k\_1^2, -k\_2^2, -k\_1k\_2$,
and the isometry group is a solvable group. If the metric were conformally flat, then
this solvable group would embed into $O(3,1)$ by [Liouville's theorem](http://en.wikipedia.org/wiki/Liouville%2527s_theorem_%28conformal_mappings%29). However, one can
check that this solvable group does not embed by analyzing the Lie algebra and
comparing it to the Lie algebras of the solvable subgroups of $O(3,1)$. 


