Finding Stationary Points of Natural Log Function $$
f(x) = x - 2\ln(x^2 + 3)
$$
I started by using the chain rule on $x^2 + 3$ which gives me
$\frac{2x}{x^2} + 3$.
At this point I tried to multiple $\frac{2x}{x^2} + 3$ by $x - 2$ - is this correct?
How can I then find the stationary point? 
I understand that you may not wish to give me the answer on a silver platter. Perhaps somebody could kindly provide me with an example in a similar format and the steps involved in solving it.

The first derivative of your function can be found like this:
$$
\begin{eqnarray}
f'(x) &=& d(x - 2 \ln(x^2 + 3))/dx \\
      &=& d(x)/dx - d(2 \ln(x^2 + 3))/dx \\
      &=& 1 - \frac{2 d(x^2 + 3)/dx}{x^2 + 3} \\
      &=& 1 - \frac{4x}{x^2 + 3} \\
\end{eqnarray}
$$
And the stationary points can be found by solving $f'(x) = 0$:
$$
\begin{eqnarray}
1 - \frac{4x}{x^2 + 3} &=& 0 \\
\frac{4x}{x^2 + 3} &=& 1 \\
x^2 + 3 &=& 4x \\
x^2 - 4x + 3 &=& 0 \\
(x - 1)(x - 3) &=& 0 \\
\end{eqnarray}
$$
So you have $x = 1$ and $x = 3$.
