Answer: The only stationary point has \(x=\sqrt2\).

The curve is

\(y=x-\ln(x^2+2x).\)

First note the domain. The logarithm requires

\(x^2+2x\gt0.\)

So

\(x(x+2)\gt0,\)

which means

\(x\gt0\quad\text{or}\quad x\lt-2.\)

Differentiate:

\(\dfrac{\mathrm dy}{\mathrm dx}=1-\dfrac{2x+2}{x^2+2x}.\)

At a stationary point, \(\dfrac{\mathrm dy}{\mathrm dx}=0\), so

\(1-\dfrac{2x+2}{x^2+2x}=0.\)

Therefore

\(x^2+2x=2x+2.\)

Subtract \(2x\) from both sides:

\(x^2=2.\)

So

\(x=\sqrt2\quad\text{or}\quad x=-\sqrt2.\)

However, \(-\sqrt2\) is between \(-2\) and \(0\), so it is not in the domain of the logarithm. The value \(\sqrt2\) is in the domain \(x\gt0\).

Therefore there is exactly one stationary point, and its \(x\)-coordinate is

\(x=\sqrt2.\)