Let $f$ be analytic in the closed annulus $A = \{z:R_1 \leq |z-z_0| \leq R_2\}$ Let $f$ be analytic in the closed annulus $A = \{z:R_1 \leq |z-z_0| \leq R_2\}$ and let $z_1$ be an interior point of $A$. Let $C_1,C_2$ be the positively oriented circles $|z-z_0| = R_1$ and $|z-z_0| = R_2$, respectively. Prove that $f(z_1) = \dfrac{1}{2\pi i}[\int_{C_2}\dfrac{f(z)}{z-z_1}dz - \int_{C_1}\dfrac{f(z)}{z-z_1}dz]$
Use the hint to solve the question: Construct two simple closed contours and apply the Cauchy's formulas accordingly.
I will attach a picture of my workings because i have drawings on my workings in which i do not know how to put in latex. Please bear with my handwriting and tell me where i went wrong and what is my misconception as i cant seem to get the right answer no matter how long i spent.
 A: You're absolutely right in how you decompose the annulus, but you made a small mistake.  $z_1$ cannot be in the interior of $D_1$ and $D_2$, that makes no sense!
But, assuming without loss of generality that $z_1$ is in $D_1$ (otherwise rotate your dividing line of the annulus), you do know the integral over $D_2$: since $z_1$ is not in the interior of $D_2$, the function $\frac{z}{z-z_1}$ is holomorphic on all of $D_2$ and you can apply the Cauchy integral theorem (which presumably you know if you are using the integral formula):  the integral of a holomorphic function around a closed path is simply $0$.  This should fix your factor of $2$ problem!
Regardless of this, there is also a mistake at the end of your work.
In the beginning you reasoned that $$f(z_1) = \frac{1}{2\pi i}\int_{D_1} \frac{f(z)}{z-z_1} = \frac{1}{2\pi i}\int_{D_2} \frac{f(z)}{z-z_1}$$
Given this, you should have found 
$$2f(z_1) = \frac{1}{2\pi i}\int_{D_1 + D_2} \frac{f(z)}{z-z_1}$$
It appears that you were simultaneously applying the properties of additivity of integration interval (correctly) and linearity of integration (erroneously).  
