= Solution
The <derivative> of an <analytic function> must be independent of the direction of approach. Taking a real increment gives $f'(z)=u_x+iv_x$; taking an imaginary increment gives
$$
f'(z)=\frac{u_y+iv_y}{i}=v_y-iu_y.
$$
Equality of real and imaginary parts proves the <Cauchy-Riemann equations>:
$$
\boxed{u_x=v_y,\qquad u_y=-v_x}.
$$
A real <harmonic function> is a twice continuously differentiable function whose <Laplacian> vanishes. For the given logarithmic function and $x^2+y^2>0$,
$$
u_x=\frac{2x}{x^2+y^2},\quad u_y=\frac{2y}{x^2+y^2},\quad
u_{xx}+u_{yy}=\frac4{x^2+y^2}-\frac{4(x^2+y^2)}{(x^2+y^2)^2}=0.
$$
It is therefore harmonic everywhere on the <punctured complex plane>.
On any domain admitting a <branch of the complex logarithm>, choose
$$
\boxed{v(x,y)=2\operatorname{Arg}(x+iy)+C,\qquad u+iv=2\operatorname{Log}z+iC}.
$$
Its <derivatives> are $v_x=-2y/(x^2+y^2)$ and $v_y=2x/(x^2+y^2)$, as required. For example, the principal <complex logarithm> gives such a <harmonic conjugate> on the plane cut along the negative real axis.
There is no single-valued <harmonic conjugate> on the whole punctured plane. Indeed its putative differential would satisfy
$$
\oint_{|z|=1}dv=\oint_{|z|=1}\left(-\frac{2y}{x^2+y^2}\,dx+\frac{2x}{x^2+y^2}\,dy\right)=4\pi,
$$
whereas a globally defined function has zero integral of its differential around a closed curve. This is the <log modulus has no global harmonic conjugate on the punctured plane> obstruction. Thus the conjugate requested in the question must be interpreted locally or on a cut plane.
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