Series expansion for $x$, when $x$ is small Suppose that we are given the series expansion of $y$ in terms of $x$, where $|x|\ll 1$. For example, consider $$y=x+x^2+x^3+\cdots\qquad\qquad\qquad (1).$$ From this I would like to derive the series expansion of $x$ in terms $y$. Given that $|x|\ll 1$ we can take $y=x+O(x^2)$ such that $x\sim y$. 
It then follows from $(1)$ that $$x=y-(x^{2}+x^{3}+\cdots)\qquad\qquad\qquad (2).$$ Given that $x\sim y$, $(2)$ becomes $$x=y-y^2-y^3+\cdots\qquad\qquad\qquad (3).$$
While the first two terms of $(3)$ are correct I suspect that the remaining terms are incorrect. 
For instance given that $y=x+O(x^2)$ implies that $y^2=x^2+2xO(x^{2})+(O(x^{2}))^{2}$. Therefore, $2xO(x^{2})$ will yield a $y^{3}$ term in $(3)$ when computing $x^{2}$ in $(2)$. 
How do I find the correct term containing $y^{3}$ in $(3)$, and so on.

This is not an answer but it is too long for a comment.
Just as Will Jagy commented, you can inverse the series building one coefficient at the time. This is a quite tedious task but it is doable.
Suppose that you have 
$$y=a_0+a_1+a_2x^2+a_3x^3+O\left(x^4\right)$$, replacing $x$ by $b_0+b_1+b_2y^2+b_3y^3$, replacing, expanding and identifying one coefficient at the time,  you will then arrive to $$x=\frac{1}{a_1}(y-a_0)-\frac{a_2 }{a_1^3}(y-a_0)^2+\frac{2 a_2^2-a_1
   a_3 }{a_1^5}(y-a_0)^3+O\left((y-a_0)^4\right)$$
If we set all the $a_i$'s equal to $1$ as in the example in the post, you will then have $$x=y-y^2+y^3+O\left(y^4\right)$$ If we replace $x$ by $y-y^2+y^3$ in $y=x+x^2+x^3$, we should get $$y=y+4y^5-6y^6+\cdots$$ which shows the error.
