= Solution
For a <holomorphic function>, the derivative supplied by <complex differentiability> must have the same value along real and imaginary increments. Those two computations give
$$
f'(z)=u_x+iv_x=\frac{u_y+iv_y}{i}=v_y-iu_y.
$$
Equating real and imaginary parts proves the <Cauchy-Riemann equations>:
$$
\boxed{u_x=v_y,\qquad u_y=-v_x.}
$$
Conversely, if the first partial <derivatives> are continuous near a point and obey these equations there, real differentiability gives
$$
f(z+h)-f(z)=(u_x+iv_x)\operatorname{Re}h+(u_y+iv_y)\operatorname{Im}h+o(|h|)=f'(z)h+o(|h|),
$$
which proves complex differentiability under this regularity assumption.
The <gradients> of the real and imaginary parts have <dot product>
$$
\nabla u\cdot\nabla v=u_xv_x+u_yv_y=v_yv_x-v_xv_y=0,
$$
and $|\nabla u|^2=|\nabla v|^2=|f'(z)|^2$. At every point with $f'(z)\ne0$, the two <level sets> are regular curves, their <gradients> are nonzero <normal vectors>, and their tangent directions are therefore perpendicular.
The printed phrase "at every point" needs this regular-point qualification. At a <critical point> with $f'(z)=0$, the <level sets> may cross or fail to have a unique tangent. For example, $f(z)=z^2$ gives $u=x^2-y^2$ and $v=2xy$: at the origin the zero <level sets> are respectively the diagonals and the coordinate axes. Thus the valid conclusion is $\boxed{\text{orthogonality at every regular intersection, where }f'(z)\ne0}$.
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