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A graph $G$ is $(d_1,d_2,d_3)$-colorable if the vertex set $V(G)$ can bepartitioned into three subsets $V_1,V_2$ and $V_3$ such that for$i\in\{1,2,3\}$, the induced graph $G[V_i]$ has maximum vertex-degree at most$d_i$. So, $(0,0,0)$-colorability is exactly 3-colorability.  The well-known Steinberg's conjecture states that every planar graph withoutcycles of length 4 or 5 is 3-colorable. As this conjecture being disproved byCohen-Addad etc. in 2017, a similar question, whether every planar graphwithout cycles of length 4 or $i$ is 3-colorable for a given $i\in\{6,\ldots,9\}$, is gaining more and more interest. In this paper, we considerthis question for the case $i=6$ from the viewpoint of improper colorings. Moreprecisely, we prove that every planar graph without cycles of length 4 or 6 is(1,0,0)-colorable, which improves on earlier results that they are(2,0,0)-colorable and also (1,1,0)-colorable, and on the result that planargraphs without cycles of length from 4 to 6 are (1,0,0)-colorable.

(1,0,0)-colorability of planar graphs without cycles of length 4 or 6