Recall that the Riemann surface of $w^3 = z$ is defined to be
$$\{(w_0, w_1, z_0, z_1) \in \Bbb R^4 \, :\, (w_0 + iw_1)^3 = z_0 + iz_1\}$$
While taking the real part you are really plotting
$$\{(x, y, z) \in \Bbb R^3 \, : \, z = \Re[(x+iy)^{1/3}]\}$$
So that's bound to be a section of the original Riemann surface, a bit deformed around the branches. Similar holds for the imaginary part.
So yes, those indeed look similar to the original Riemann surface.