If you "connect" the original path $\,|z|=4\,$ with those two paths $\,|z-2|=1/2\,,\,|z-3|=1/2\,$ via straight lines and you integrate forth and back those straight segments and around the new paths, the integral over each straight segment cancels out (both directions!) and you remain only with the integrals over the new paths, outside of which the integral equals zero as $\,\frac{1}{(z-2)(z-3)}\,$ is analytic.
Thus, the integral equals the sum of $\oint_{|z-2|=1/2}\frac{dz/(z-3)}{z-2}=2\pi i\left.\left(\frac{1}{z-3}\right)\right|_{z=2}=-2\pi i$ $\oint_{|z-3|=1/2}\frac{dz/(z-2)}{z-3}=2\pi i\left.\left(\frac{1}{z-2}\right)\right|_{z=3}=2\pi i$ All in all, we get that $\oint_{|z|=4}\frac{dz}{(z-2)(z-3)}=-2\pi i+2\pi i =0$