Help with dy/dx of natural log 
Could someone explain how I would work this problem with steps? 

It is easier to simplify first:
$$
y = \ln x - \ln(1+x^2).
$$
Then
\begin{align*}
\frac{dy}{dx} & = \frac{d}{dx}\ln x - \frac{d}{dx} \ln(1+x^2).
\end{align*}
The first derivative is $1/x$. For the second derivative, we need to use the chain rule:
\begin{align*}
\frac{d}{dx} \ln(1+x^2) & = \frac{d\ln(1+x^2)}{d(1+x^2)} \cdot \frac{d(1+x^2)}{dx} \\
& = \frac{1}{1+x^2} \cdot 2x = \frac{2x}{1+x^2}.
\end{align*}
Hence 
$$
\frac{dy}{dx} = \frac{1}{x}-\frac{2x}{1+x^2}.
$$
