Step 1: Work in the rest frame of the decaying particle. Its four-momentum is purely temporal, $E_{\text{tot}} = mc^2$ and $\vec{p}_{\text{tot}} = 0$. Momentum conservation forces the two fragments to be back-to-back with identical magnitude of momentum, so they share the same speed $v$.
Step 2: Energy conservation with rest mass $m' = \tfrac{3}{10}m$ for each fragment gives $mc^2 = 2\gamma m'c^2$, i.e. $\gamma = \dfrac{m}{2m'} = \dfrac{m}{2\cdot 0.3m} = \dfrac{1}{0.6} = \dfrac{5}{3}$.
Step 3: Use $\beta = \sqrt{1 - 1/\gamma^2}$ where $\beta = v/c$. With $\gamma = 5/3$, $\;1/\gamma^2 = 9/25$, so \[\beta = \sqrt{1 - \tfrac{9}{25}} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5} = 0.8.\]
Step 4: Thus each fragment moves with speed $0.8c$ along the z-axis, one in $+\hat{z}$ and the other in $-\hat{z}$. Rest-mass energy is not conserved as such; the missing rest energy $mc^2 - 2m'c^2 = 0.4mc^2$ becomes the kinetic energy of the fragments.\[\boxed{v_1 = -v_2 = 0.8c\,\hat{z}}\]