Find the derivative of $\frac{2}{x+1}-\frac{x^{2}}{3x-1}$.

Let $f(x) = \frac{2}{x+1} - \frac{x^2}{3x-1}$.
Using the linearity of the derivative:
$f'(x) = \frac{d}{dx}\left(\frac{2}{x+1}\right) - \frac{d}{dx}\left(\frac{x^2}{3x-1}\right)$.
Applying the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2}$:
For the first term: $\frac{d}{dx}\left(\frac{2}{x+1}\right) = \frac{(x+1)(0) - 2(1)}{(x+1)^2} = \frac{-2}{(x+1)^2}$.
For the second term: $\frac{d}{dx}\left(\frac{x^2}{3x-1}\right) = \frac{(3x-1)(2x) - x^2(3)}{(3x-1)^2} = \frac{6x^2 - 2x - 3x^2}{(3x-1)^2} = \frac{3x^2 - 2x}{(3x-1)^2}$.
Combining these results:
$f'(x) = -\frac{2}{(x+1)^2} - \frac{3x^2 - 2x}{(3x-1)^2}$.