f(-x) = -((1+x)/(1-x))

Given $f(x) = \ln\left(\frac{1+x}{1-x}\right)$.
Using the logarithm property $\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$:

$$f(x) = \ln(1+x) - \ln(1-x)$$

Differentiate $f(x)$ with respect to $x$. Recall that $\frac{d}{dx}(\ln(u)) = \frac{1}{u} \frac{du}{dx}$.
For the first term, $\frac{d}{dx}(\ln(1+x)) = \frac{1}{1+x} \cdot \frac{d}{dx}(1+x) = \frac{1}{1+x} \cdot 1 = \frac{1}{1+x}$.
For the second term, $\frac{d}{dx}(\ln(1-x)) = \frac{1}{1-x} \cdot \frac{d}{dx}(1-x) = \frac{1}{1-x} \cdot (-1) = -\frac{1}{1-x}$.
So, $f'(x)$ is:

$$f'(x) = \frac{1}{1+x} - \left(-\frac{1}{1-x}\right)$$ $$f'(x) = \frac{1}{1+x} + \frac{1}{1-x}$$

Find a common denominator, which is $(1+x)(1-x)$:

$$f'(x) = \frac{1(1-x)}{(1+x)(1-x)} + \frac{1(1+x)}{(1-x)(1+x)}$$ $$f'(x) = \frac{1-x+1+x}{(1+x)(1-x)}$$ $$f'(x) = \frac{2}{1-x^2}$$

The expression for $f'(x)$ as a single simplified fraction is $\boxed{\frac{2}{1-x^2}}$.