Let $\ell$ be tangent to the hyperbola at point $B$, $F$ be the intersection of $x$-axis and $BC$, $E$ be the intersection point of the $x$-axis and $\ell$,

$D$ be the intersection point of $\ell$ and the line through $A$ parallel to $y$-axis (see the Fig.). Since $A$ and $B$ are symmetric with respect to the origin, we see that $A(-x_1; -1/x_1)$. Further,

$$\tan \angle \text{BEO} = \tan(\pi - \angle \text{BEF}) = -\tan \angle \text{BEF} = - (1/x)' \bigg|_{x=x_1} = \frac{1}{x_1^2}.$$

On the other hand, $\tan \angle BAD = \tan \angle BOE = 1/x_1 : x_1 = 1/x_1^2$, i.e.
$$\angle BAD = \angle BEO. \tag{1}$$

Further,
$$\tan \angle BFE = \tan \angle BCG = \frac{1/x_1 - 1/x_2}{x_2 - x_1} = \frac{1}{x_1 x_2}$$
and
$$\tan \angle CAD = \frac{1/x_2 + 1/x_1}{x_2 + x_1} = \frac{1}{x_1 x_2}.$$
Hence $\angle CAD = \angle CFE = \angle BFE$.

So,
$$\angle BAC = \angle BAD - \angle CAD \stackrel{(1)}{=} \angle BEO - \angle BFE = \angle FBE = \angle CBE,$$

as required.