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It seems to me that the answer is NO. Namely, if $(M,g)$ is compact, conformally flat and has zero scalar curvature then $g$ is a flat metric. Indeed, by the hypothesis there is a smooth function $f:M \to \mathbb R$ such that $e^f g$ is a flat metric on $M$. Then a finite covering of $M$ is a torus. The metric $g$ then lift to such a torus and by the Gromov-Lawson theorem (see Corollary A in http://www.ihes.fr/~gromov/PDF/8%5B26%5D.pdf) the lifted metric is flat hence the original $g$ is flat. QED