We have $f(z)dz=f(x+iy)d(x+iy)=(u+iv)(dx+idy)=udx+iudy+ivdx-vdy$ Thus we can write the contour integral as the line integral of a differential form
$\displaystyle\oint_{C(z_0,r)}f(z)dz=\displaystyle\oint_{C(z_0,r)}udx-vdy+i\displaystyle\oint_{C(z_0,r)}vdx+udy$
Since $u,v$ are both differentiable by assumption, we can apply Green's Theorem and get
$\displaystyle\oint_{C(z_0,r)}f(z)dz=\displaystyle\iint_{D(z_0,r)}(-\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y})dxdy+i\displaystyle\iint_{D(z_0,r)}(-\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y})dxdy$
Now apply mean value thm for double integral
$\displaystyle\oint_{C(z_0,r)}f(z)dz=\pi r^2[-\frac{\partial v}{\partial x}(z_0+\rho e^{i\theta})-\frac{\partial u}{\partial y}(z_0+\rho e^{i\theta})]+i\pi r^2 [-\frac{\partial u}{\partial x}(z_0+\rho' e^{i\theta'})-\frac{\partial v}{\partial y}(z_0+\rho' e^{i\theta'})]$
for suitable $0<\rho,\rho' and $0<\theta,\theta'<2\pi$
As $r$ tends to $0^{+}$, $z_0+\rho e^{i\theta}$ and $z_0+\rho' e^{i\theta'}$ tend to $z_0$, uniformly w.r.t $\theta$ and $\theta'$.
Hence the RHS of last equation (divided by $\pi r^2$)goes to zero iff $u,v$ satisfy CR-equation in $z_0$, i.e iff $f$ is differentiable in the complex sense at $z_0$