If $$\int \frac{dx}{(x^2 - 2x + 10)^2} = A\left(\tan^{-1}\left(\frac{x-1}{3}\right) + \frac{f(x)}{x^2 - 2x + 10}\right) + C$$, then (where C is a constant of integration)

We begin with the given integral

$$I=\int \frac{dx}{(x^{2}-2x+10)^{2}}.$$

First, we complete the square in the quadratic expression that appears in the denominator. We have

$$x^{2}-2x+10=x^{2}-2x+1+9=(x-1)^{2}+3^{2}.$$

So the integral becomes

$$I=\int \frac{dx}{\bigl((x-1)^{2}+3^{2}\bigr)^{2}}.$$

To simplify the expression, we introduce a new variable

$$u=x-1 \quad\Longrightarrow\quad du=dx.$$

Substituting $$u$$ for $$(x-1)$$ and $$du$$ for $$dx$$ gives

$$I=\int \frac{du}{(u^{2}+3^{2})^{2}}=\int \frac{du}{(u^{2}+9)^{2}}.$$

Next, we use the standard formula for the integral of the reciprocal of a squared quadratic:

Formula. For any positive constant $$a$$,

$$\int \frac{du}{(u^{2}+a^{2})^{2}} =\frac{u}{2a^{2}\,(u^{2}+a^{2})}+\frac{1}{2a^{3}}\tan^{-1}\!\Bigl(\frac{u}{a}\Bigr)+C.$$

Here $$a=3$$, so $$a^{2}=9$$ and $$a^{3}=27$$. Substituting $$a=3$$ into the formula gives

$$I=\frac{u}{2\cdot 9\,(u^{2}+9)}+\frac{1}{2\cdot 27}\tan^{-1}\!\Bigl(\frac{u}{3}\Bigr)+C =\frac{u}{18\,(u^{2}+9)}+\frac{1}{54}\tan^{-1}\!\Bigl(\frac{u}{3}\Bigr)+C.$$

We now return to the original variable by writing $$u=x-1$$ and $$u^{2}+9=(x-1)^{2}+9=x^{2}-2x+10$$. Thus

$$I=\frac{x-1}{18\,(x^{2}-2x+10)}+\frac{1}{54}\tan^{-1}\!\Bigl(\frac{x-1}{3}\Bigr)+C.$$

To express this result in the exact form required, we extract the common factor $$\dfrac{1}{54}$$:

$$I=\frac{1}{54}\left[\tan^{-1}\!\Bigl(\frac{x-1}{3}\Bigr) +\frac{3(x-1)}{x^{2}-2x+10}\right]+C.$$

Comparing with the structure

$$I=A\left(\tan^{-1}\!\Bigl(\frac{x-1}{3}\Bigr)+\frac{f(x)}{x^{2}-2x+10}\right)+C,$$

we identify

$$A=\frac{1}{54}\qquad\text{and}\qquad f(x)=3(x-1).$$

These values correspond precisely to Option D.

Hence, the correct answer is Option D.