= Solution
At $z_0$, <complex differentiability> gives
$$
f(z_0+h)=f(z_0)+f'(z_0)h+o(|h|).
$$
Writing $f'(z_0)=\lambda e^{i\alpha}$, the assumption $f'(z_0)\ne0$ gives $\lambda>0$. Its real derivative is therefore the similarity $h\mapsto\lambda e^{i\alpha}h$, a rotation through $\alpha$ followed by scaling by $\lambda$.
More explicitly, for two differentiable curves $\gamma_1,\gamma_2$ with $\gamma_j(0)=z_0$ and nonzero tangent vectors, the <chain rule> gives
$$
(f\circ\gamma_j)'(0)=f'(z_0)\gamma_j'(0),\qquad
\frac{(f\circ\gamma_2)'(0)}{(f\circ\gamma_1)'(0)}=\frac{\gamma_2'(0)}{\gamma_1'(0)}.
$$
Taking the <complex argument> of this ratio shows that the oriented angle is preserved, while nonvanishing ensures both image tangents remain regular. Thus
$$
\boxed{f\text{ holomorphic and }f'\ne0\text{ on }U\ \Longrightarrow\ f\text{ conformal on }U}.
$$
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