This decay (occurring via the strong interaction) violates the charge conjugation since $J^{PC}(\pi^0) = 0^{-+}, J^{PC}(\rho^0) = 1^{--}, J^{PC}(\eta'^0) = 0^{-+}$.
The charge conjugation transforms a particle in its anti-particle. In the case of the 3 particles involved in this decay, they are all their own anti-particle, and the effect of the charge conjugation operator $C$ is therefore (taking as an example the pion) $C|\pi^0> = \eta_C |\pi^0>$, meaning that the $\pi^0$ is eigenstate of the charge conjugation with eigenvalue $\eta_C = +1$. The $\rho^0$ has $\eta_C=-1$ and the $\eta'^0$, +1 (remark: $\eta_C$ is necessarily $\pm 1$ because when you apply twice the charge conjugation you should recover the initial state). The requirement of the charge conjugation conservation by the strong interaction would imposes: $\eta_C(\eta'^0) = \eta_C(\pi^0) \times \eta_C(\rho^0)$ which is not the case $+1 \ne (+1) \times (-1)$. Thus this reaction is forbidden.