This is very hard. Does anyone have a solution? An easier version is when we cancel the word "positive". Then we can solve with the identity $ \left(\frac{-x^3+3x+1}{x^2+x+1}\right)^3 + \left(\frac{3x^2+3x}{x^2+x+1}\right)^3 + (x-1)^3 = 9x$. For the given version, I have no idea, can anyone type a proof here?

It's easy using your identity: To represent some $ y>0$, it suffices to represent some $ yr^3$ for at least one $ r>0$. Now choose $ r$ such that $ x=\frac{yr^3}9$ has the property that all three terms on your RHS are positive (for example if $ x \in ]1, \frac 32]$). $ r$ can be choosen as the cubes are dense. Thus $ 9x = yr^3$ is a sum of three positive cubes, thus so is $ y$.

Hmm... Good point. Didn't see that. Weird, I vaguely remember seeing a hard proof with complex analysis, probably I'm confusing things then.