I would like to verify the following statement. However, I do not know what should I do. Generally, I think that, since there is \$1/2\pi i$ part, Winding number, Residue theorem and Argument Principle might be helpful. But the condition required in those theorem does not exactly meet the given condition of this problem. :
Suppose that $f$ is holomorphic in neighborhood of the closure of the unit disc. If $|z| \leq 1$, we have $$f(z)(1- |z|^2) = \frac{1}{2\pi i}\int_{|w| = 1} \frac{1 -\bar zw }{w-z}f(w)\ dw.$$
For a circle $C$ centered on the origin of radius $r$ and a function $f$ holomorphic on and within $C$, we have Cauchy's Integral Formula for $z$ in the interior of $C$:
$$ f(z) = \frac{1}{2\pi i} \int_C \frac{f(w)}{w-z}dw$$
Already at this point we recognise the first half of the desired expression.
Now consider the function $ g(z) = z f(z)$; $g$ is holomorphic everywhere $f$ is, so we may use Cauchy's Integral Formula again:
$$ zf(z) = g(z) = \frac{1}{2\pi i} \int_C \frac{g(w)}{w-z}dw = \frac{1}{2\pi i} \int_C \frac{wf(w)}{w-z}dw$$
So
$$f(z) = \frac{1}{2\pi i z} \int_C \frac{wf(w)}{w-z}dw = \frac{1}{2\pi i |z|^2} \int_C \frac{\bar zwf(w)}{w-z}dw $$
And the desired expression follows. This is true for $|z| < 1$.
All that remains is to show that the expression holds for $|z| = 1$. Under this assumption, $f(z)(1 - |z|^2) = 0$, so it suffices to show the right hand side vanishes also. Consider the following:
$$\begin{align} \frac{1}{\bar z} \int_{|w| = 1} \frac{1 -\bar zw }{w-z}f(w)\ dw & = \int_{|w| = 1} \frac{1 -\bar zw }{\bar zw-\bar zz}f(w)\ dw \\ & = \int_{|w| = 1} \frac{1 -\bar zw }{\bar zw-1}f(w)\ dw \\ & = -\int_{|w| = 1} f(w) dw = 0 , \, \text{by Cauchy's Theorem} \end{align}$$
So since $\bar z$ is non-zero, the expression holds for all $z$ on and within the unit disc.