Does there exist an Einstein manifold which is not conformally flat, which is to say one which has nonvanishing Weyl tensor. If so, what is a good example.

All conformally flat homogeneous riemannian manifolds are symmetric spaces, by a result of Takagi. All homogeneous riemannian manifolds of dimension $\leq 11$ admit Einstein metrics, by results of Wang and Ziller. Most homogensous riemannian manifolds are not symmetric spaces. References:Takagi, H.: Conformally flat Riemannian manifolds admitting a transitive group of isometries I, II. Tohoku Math. J. 27, 103–110(I), 445–451(II) (1975) McKenzie Y. Wang, Wolfgang Ziller: Existence and nonexistence of homogeneous Einstein metrics, Inventiones mathematicae, 1986, Volume 84, Issue 1, pp 177194 


K3 surface is Einstein, by CalabiYau theorem, but it is not conformally flat; this can be seen e.g. from topology, or from the reference that José FigueroaO'Farrill has given above. 


Does your question refer to both Riemannian and semiRiemannian spaces? If so, then the Schwarzschild metric is an example. It's a vacuum solution to the Einstein field equations (with zero cosmological constant), so it's an Einstein manifold. It has a nonvanishing Weyl tensor. 


By the standard curvature decomposition, for an Einstein metric, if the Weyl curvature vanishes, then the sectional curvature is constant, so other than space forms, all Einstein manifolds are not conformally flat. Sorry I didn't see Deane Yang's answer, he already got this answer. 

