Mistake proving that $3 = 0$

Consider the equation $x=x^2$. Its solutions are easy to find: they are $0$ and $1$, right?!

However, if $x$ is a solution of that equation, then $x=x^2$, and therefore $x=(x^2)^2=x^4$. But the equation $x=x^4$ has other solutions, besides $0$ and $1$, namely $-\frac12+\frac{\sqrt3}2i$.

The problem lies in the if-then clause of the first sentence of the previous paragraph: we are saying (correctly) that if $x$ is a solution of the given equation, then $x$ is also a solution of another equation. But there is no reason for you to think that the new equation only has the solutions of the original one!

That's what happens in your example: if $x$ is a solution of $x^2+x+1=0$, then $x$ is a solution of $x^2=\frac1x$, but you have no reason to suppose that every solution of $x^2=\frac1x$ is also a solution of $x^2+x+1=0$.

The reasonning only shows that if $x$ satisfies $x^2 + x+1=0$ then it satisfies $x^3=1$. The fact is that you cannot "go back" to prove the reverse, since at step 3 you reuse the hypothesis that $x^2+x+1=0$.