The angle between two curves through a point is measured by means of the tangent vectors to the curves at the point. Then, a map preserves angles (i.e., it is conformal) if the differential of the map preserves angles. The differential of the map is the linear map defined by the Jacobian matrix. As you have said, the Jacobian corresponds to a conformal linear map, thus proving that holomorphic functions are conformal.
In the case of manifolds, a map f between two manifolds M and M’ induces a linear map between the tangent space of M at p and the tangent space of M’ at f(p), which is called the differential of f at p. The map is conformal if the differential is conformal at every point. In the present case both manifolds M and M’ are the set of complex numbers.
Write
$$f(x,y)=u(x,y)+i v(x,y),\quad g(\xi,\eta)=a(\xi,\eta)+ib(\xi,\eta)$$
where $u$, $v$, $a$, $b$ are realvalued functions defined in $A$, resp. $B$. Then by definition of $g$ one has
$$a(\xi,\eta)+ib(\xi,\eta)=g(\xi+i\eta)=\overline{f(\xi-i\eta)}=u(\xi,-\eta)-iv(\xi,-\eta)$$
and therefore
$$a(\xi,\eta)=u(\xi,-\eta),\quad b(\xi,\eta)=-v(\xi,-\eta)\qquad\bigl((\xi,\eta)\in B\bigr)\ .$$
It follows that, e.g. $$b_\eta(\xi,\eta)=-v_y(\xi,-\eta)\cdot(-1)=v_y(\xi,-\eta)\ .$$
Since $u$ and $v$ satisfy the CR-equations in the variables $x$ and $y$ we conclude that
$$a_\xi(\xi,\eta)=u_x(\xi,-\eta)=v_y(\xi,-\eta)=b_\eta(\xi,\eta)\ ,$$
and similarly
$$a_\eta(\xi,\eta)=-u_y(\xi,-\eta)=v_x(\xi,-\eta)=-b_\xi(\xi,\eta)\ .$$
This shows that $g$ fulfills the CR-equations in the variables $\xi$ and $\eta$.
But there is also a direct approach, which in my view is simpler and more in tune with a complex world description.
As $f$ is holomorphic in $A$, for each point $z_0\in A$ (held fixed in the following) there is a complex number $C$ such that
$$f(z)-f(z_0)=C(z-z_0)+o(|z-z_0|)\qquad (z\in A, \ z\to z_0)\ .$$
Let a point $w_0\in B$ be given, and put $z_0:=\bar w_0$. Then by definition of $g$ one has
$$g(w)-g(w_0)=\overline{f(\bar w)}-\overline{f(\bar w_0)}=\overline{f(\bar w)-f( z_0)}=\overline{C(\bar w -z_0)+o(|\bar w-z_0|)}\qquad(w\in B)\ .$$
As $|\bar w -z_0|=|w-w_0|$ it follows that
$$g(w)-g(w_0)=\bar C(w-w_0)+o(|w-w_0|)\qquad(w\in B, \ w\to w_0)\ .$$
It follows that $g'(w_0)=\bar C$, and as $w_0\in B$ was arbitrary, we conclude that $g$ is holomorphic in $B$.
Best Answer
Hint: In general, use these substitutions \begin{align} x=\frac{z+\bar{z}}{2}\\ y=\frac{z-\bar{z}}{2i}\\ |z|^2=x^2+y^2\\ =z\bar{z} \end{align} Edit: After substitution \begin{align} u &= \dfrac{\frac{z+\bar{z}}{2}\left(1+z\overline{z}\right)}{1+z^2+\overline{z}^2+z^2\overline{z}^2}\\ &= \dfrac12\dfrac{z(1+\overline{z}^2)+\overline{z}(1+z^2)}{(1+z^2)(1+\overline{z}^2)}\\ &= \dfrac12\left(\dfrac{z}{1+z^2}+\dfrac{\overline{z}}{1+\overline{z}^2}\right) \end{align} then $$v=\dfrac{1}{2i}\left(\dfrac{z}{1+z^2}-\dfrac{\overline{z}}{1+\overline{z}^2}\right)$$ and $$\color{blue}{f(z)=\dfrac{z}{1+z^2}}$$