= Solution
For $f=u+iv$ holomorphic in $z=x+iy$, the <Cauchy-Riemann equations> are $u_x=v_y$ and $u_y=-v_x$. Therefore the normal <vectors> to the two level curves obey
$$
\nabla u\cdot\nabla v=u_xv_x+u_yv_y=u_x(-u_y)+u_yu_x=0.
$$
At a point where $f'\ne0$, both <gradients> are nonzero, so their perpendicular tangent lines are also perpendicular. At critical points this argument does not assert that regular level curves exist.
Take
$$
\boxed{f(z)=z^2.}
$$
Its <real part> is $u=x^2-y^2$, so the specified first family is $u=-\alpha$. Its <imaginary part> is $v=2xy$, so the second family is $v=2\beta$. Since $\beta\ne0$, their intersections exclude the origin, where $f'(z)=2z$ would vanish. Thus these <hyperbolas> meet orthogonally as required.
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