Are negatively pinched manifold locally conformally flat? 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?
 A: Another easy locally homogeneous counterexample to the original question is given by complex hyperbolic manifolds. This includes compact examples. Complex hyperbolic manifolds are Einstein. Curvature tensor of any manifold decomposes into its Weyl part+Ricci part +scalar part. Thus, a conformally flat Einstein manifold must necessarily have scalar curvature operator and hence have constant sectional curvature in dimensions above 2. This is definitely not the case for complex hyperbolic manifolds so they are not locally conformally flat. Also, as Igor mentioned Chern-Weil theory in dimension 4 says that $sig(M^4)=\frac{1}{12\pi^2}\int_M(|W^+|^2-|W^-|^2)$, where $W^\pm$ are self-dual and anti-self-dual parts of $W$.
4-dimensional complex hyperbolic manifolds are conformally semi-flat (i.e they have $W^-=0$) which can be easily derived from the fact that their curvature tensors are $U(2)$ invariant. Thus, for a closed complex hyperbolic 4-manifold its signature is $sig(M^4)=\frac{1}{12\pi^2}\int_M(|W^+|^2\ne 0$. Moreover, the integrant is just a constant (by homogeneity). On the other hand, locally conformally flat closed 4-manifolds have signature 0 by the above formula.
A: No in dimensions $\geq 3$. To be conformally flat in 3 dimensions, the Cotton tensor must to vanish, and in dimensions $\geq 4$, the 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. 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 $O(3,1)$. 
A: Regarding vanishing rational Pontryagin classes (which do vanish for conformally flat manifolds):

*

*Recent result of Ontaneda gives examples in each dimension $\ge 4$ of closed manifolds with nonzero rational Pontryagin classes that are pinched arbitrary close to $-1$, see Corollary 4 of his paper "Pinched smooth hyperbolization" [arXiv:1110.6374].


*On the other hand, if you restrict topology of your negatively pinched $n$-manifolds in a suitable way, then one can prove vanishing of Pontryagin classes for pinching close enough to $-1$. For example, for closed manifolds of uniformly bounded simplicial volume,
if the pinching is close enough to $-1$, the manifold is diffeomorphic to a hyperbolic one (this is due to Gromov), and hence has zero Pontryagin classes. Long ago I proved similar results in the noncompact case, e.g. if you fix the fundamental group, and the dimension,
and assume the metric is complete and the fundamental group is hyperbolic, then the Pontryagin classes vanish for pinching close to $-1$, see my paper
Pinching, Pontrjagin classes, and negatively curved vector bundles. (I should mention that
my proof depends on an accessibility result of Delzant-Potyagailo in which a gap was recently discovered by Louder-Touikan who announced a fix for hyperbolic groups, see
 Larsen Louder's Research Statement. Without the fix I can only handle the case of hyperbolic groups that do not split over {1} ot $\mathbb Z$.).
