If `f'(x)=1/(3-x)^2`, the equation of the tangent line to f(x) at `(0, 1/3)` is `y=x/9+1/3`.

Explanation

To find the equation of the tangent line to the function \( f(x) \) at the point \( (0, \frac{1}{3}) \), given that \( f'(x) = \frac{1}{(3-x)^2} \), we can follow these steps:

### Step 1: Find the slope of the tangent line at the point \( x = 0 \)

We start by evaluating the derivative \( f'(x) \) at \( x = 0 \):

\[
f'(0) = \frac{1}{(3-0)^2} = \frac{1}{3^2} = \frac{1}{9}
...