<s>[INST] Are all integers not congruent to 6 modulo 9 of the form $x^3+y^3+3z^2$? Are all integers not congruent to 6 modulo 9 of the form $x^3+y^3+3z^2$
for possibly negative integers $x,y,z$?
We have the identity $ (-t)^3+(t-1)^3+3 t^2=3t-1$.
The only congruence obstruction we found is 6 modulo 9.
 [/INST] We found another approach based on integral points on genus 0 curves.
Let $G(x,z,a_1,a_2,a_3)=(a_1 x+a_2)^3+(-a_1 x+a_3)^3+3 z^2$.
For fixed $n$ and $a_i$, $G=n$ is degree two genus 0 curve and it might
have infinitely many integral points, which gives infinitely many solutions
to the OP.
WolramAlpna on solve (10x-5)^3+(-10x+3)^3+3*y^2=25 over integers.
Related paper is Representation of an Integer as a Sum of Four Integer Cubes and Related Problems
Added 22 Dec 2020
We can solve $n \equiv 1 \pmod{3}$
The solutions of $(-x-1)^3+(x+2)^3 - 3y^2=(3N+1)$
are $x=-1/2\sqrt{4y^2 + 4N + 1} - 3/2$.
When we express $4N+1$ as difference of two squares $4N+1=z^2-y'^2$ with $y'$
even we have solution.
Currently the only unsolved case appears to be $n \equiv 0 \pmod{9}$.
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