= Solution
Let $v_1,v_2\ne0$ be tangent vectors to the two smooth curves at $p$. <complex differentiability at a point> gives
$$
f(p+h)=f(p)+f'(p)h+o(|h|).
$$
When $f'(p)\ne0$, its real derivative is multiplication by a nonzero complex number, which is a rotation followed by a positive scaling. It therefore preserves the angle between $v_1$ and $v_2$, so $f$ is <conformal map>[conformal] at $p$. The condition is essential: $f(z)=z^2$ has $f'(0)=0$ and sends rays making angle $\pi/4$ at zero to rays making angle $\pi/2$.
For $J(z)=z+z^{-1}$,
$$
J(z)=J(w)\quad\Longleftrightarrow\quad (z-w)(zw-1)=0.
$$
If $z,w$ both satisfy $|z|>1$, or both satisfy $0<|z|<1$, the alternative $zw=1$ is impossible unless the points lie on the omitted unit circle. Hence $J$ is one-to-one on each region. Also $J'(z)=1-z^{-2}$ has no zero there, so the restrictions are conformal. Solving $z^2-wz+1=0$ shows that $w\in[-2,2]$ exactly when both roots lie on the unit circle; otherwise one root is inside and the other outside. Thus each region has image
$$
\boxed{\mathbb C\setminus[-2,2]}.
$$
Taking the reciprocal of the interior restriction and filling its removable value at zero gives
$$
\boxed{g(z)=\frac1{J(z)}=\frac{z}{1+z^2}}.
$$
The reciprocal sends $\mathbb C\setminus[-2,2]$ onto
$$
\mathbb C\setminus\bigl(({-\infty},-1/2]\cup[1/2,\infty)\bigr),
$$
and $g(0)=0$. Therefore $g$ is the required one-to-one conformal map from the <unit disc>.
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