I do not know if I did this fully correctly this is what I have:
$$f(z) = (x + iy)^2 - y^2 = (x^2 + 2xyi - y^2) - y^2$$
so $u(x,y) = x^2 - 2y^2$, $v(x,y) = 2xy$
so by C-R
$$\frac{\partial u}{\partial x} = 2x,\quad \frac{\partial u}{\partial y}= -4y,\quad \frac{\partial v}{\partial x}= 2y,\quad \frac{\partial v}{\partial y}= 2x$$
$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ stand but $\frac{\partial v}{\partial x}\neq \frac{\partial u}{\partial y}$ thus this is not analytic.
Is this correct or did I do this wrong?
Your line of reasoning is correct: the function $f(z)=z^2-y^2$ is not analytic. In $\mathbb{C}$ a function is analytic iff it's holomorphic: a necessary condition for that is that the function $f$ satisfies Cauchy-Riemann equations. As you noted, in our case we $f$ does not satisfies them and it is thus not analytic.
Another way to prove this is to directly prove that the function is not holomorphic:
$\begin{align}f'(1)=\lim_{z\to 1}\frac{f(z)-f(1)}{z}= \lim_{(x,y)\to (1,1)}\frac{x^2-2y^2+i2xy-(1-2+2i)}{x+iy}=\\= \begin{cases} \lim_{(x,1)\to (1,1)}\frac{x^2-2+i2x+1-2i}{x-1}=2i\\ \lim_{(1,y)\to (1,1)}\frac{1-2y^2+i2y+1-2i}{iy-i}=2\end{cases}\end{align}$
The limit does not exists, so the function is not derivable, and it follows that it is not analytic.
As a rule of thumb, most of the functions that don't have any simmetry between the variables $x,y$ are not analytic.