Calculate $\sqrt{2i}$ I did:
$\sqrt{2i} = x+yi \Leftrightarrow i = \frac{(x+yi)^2}{2} \Leftrightarrow i = \frac{x^2+2xyi+(yi)^2}{2} \Leftrightarrow i = \frac{x^2-y^2+2xyi}{2} \Leftrightarrow \frac{x^2-y^2}{2} = 0 \land \frac{2xy}{2} = 1$
$$\begin{cases}
     \frac{x^2-y^2}{2} = 0  \\
     xy = 1\\   \end{cases}   \\
=\begin{cases}
     x^2-y^2 = 0  \\
     x = \frac{1}{y}\\   \end{cases}   \\
=\begin{cases}
     \frac{1}{y}-y^2 = 0  \\
     x = \frac{1}{y}\\   \end{cases}   \\=
\begin{cases}
     \frac{1-y^3}{y} = 0  \\
     -\\   \end{cases}   
\\=
\begin{cases}
     y^3 = 1  \\
     -\\   \end{cases}  
\\=
\begin{cases}
     y = 1  \\
     x =1\\   \end{cases}    $$
And so $\sqrt{2i} = 1+i$, but my book states the solution is $\sqrt{2i} = 1+i$ and $\sqrt{2i} = -1-i$.
What did I forget?

I'll stick with your solving algorithm. You obtained (apart from a mistake):
$$\frac{1}{y^2}-y^2 = 0, \implies \frac{1}{y^2} = y^2$$
You can check that both $1$ and $-1$ satisfy this equation! So your chain of systems splits in two: knowing $x = \frac{1}{y}$,
$$y=1 \implies x = 1 \qquad y=-1 \implies x= -1 $$
So you get (as one should expect according to the Fundamental Theorem of Algebra) the two solutions
$$\sqrt{2i} = \pm(1+i) $$
Of course, you could have avoided this lengthy procedure in a few ways, for example by taking advantage of the polar form of $2i$. Take a peek at the other answers!
