solve the equation in $\Bbb C $

An easier solution than expanding like you did:

$$x^2+\left(\frac{x}{x+1}\right)^2=3$$

$$\left(x-\frac{x}{x+1}\right)^2+\frac{2x^2}{x+1}=3$$

Put $\frac{x^2}{x+1}=t$. Then, $t^2+2t=3$. So, $t=1$ or $t=-3$.

This gives that either $x^2-x-1=0$ or $x^2+3x+3=0$ which can be solved easily.

A strange solution that uses symmetry reasoning to transform the problem.

The argument $\frac{x}{1+x}$ stick out and can be set as a new variable. In fact, let's throw in a minus sign for good measure (especially if you recognize it as one of the functions that is inverse of itself).

$$u=-\frac{x}{1+x}\quad \Rightarrow \quad x=-\frac{u}{1+u}$$

If you put this in, you get the same equation for $u$, meaning that both $x$ and $u$ are solutions of this equation. Therefore, instead of one fourth order equation, we have two completely symmetric equations for two variables:

$$\begin{align}x^2+u^2&=3 \\ xu+u+x&=0\quad (\text{from }u=-\frac{x}{1+x}) \end{align} $$

Adding twice the second equation to the first one gives us $$(x+u)^2+2(x+u)-3=0$$ which is factorized by hand: $$\boxed{x+u=\{-3,1\}}$$

Subtracting twice the second equation gives

$$(x-u)^2=3+2(u+x)$$ and using the just derived solutions, $$\boxed{x-u=\pm\sqrt{\{-3,5\}}}$$

Averaging the two boxed results gives you all four solutions:

$$x=\frac{-3\pm i\sqrt{3}}{2}$$ $$x=\frac{1\pm \sqrt{5}}{2}$$

Note that computing $u$ must give you the same results (that was the premise of this method), and indeed, the $\pm$ in the second boxed equation ensures that swapping the sign just gives you the associated $u$ to each $x$.