= Solution
Let two regular differentiable paths through $z_0$ have nonzero tangent vectors $v_1,v_2$. The chain rule gives tangent vectors $f'(z_0)v_1$ and $f'(z_0)v_2$ for their images. Multiplication by a nonzero complex number is a positive scaling and <rotation>: writing $f'(z_0)=re^{i\theta}$ scales lengths by $r>0$ and adds the same angle $\theta$ to both tangents. Equivalently,
$$
\arg\frac{f'(z_0)v_2}{f'(z_0)v_1}=\arg\frac{v_2}{v_1}.
$$
Thus the oriented angle between the paths is preserved, proving \b[$f$ is conformal at $z_0$]. The nonzero derivative also gives local invertibility; a zero derivative could multiply angles through branching instead.
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