How to find the second derivative of y in $y^2 = x^2 + 2x$? I have a problem to solve:

use implicit differentiation to find $\frac{dy}{dx}$ and then $\frac{d^2y}{dx^2}$. Write the solutions in terms of x and y only

It means that I need to differentiate the equation one time to find $y'$ and then once more to find $y''$.
The correct answer from the textbook is $y' = \frac{x + 1}{y}$ and $y'' = \frac{x^2 + 2x}{y^3}$. I got the first derivative right, but I can't understand how did they get the second one, or is it a typo (unlikely), since I have $y'' = \frac{1}{y} - \frac{(x + 1)^2}{y^3}$
I did this:
$$
y^2 = x^2 + 2x\\
2yy' = 2x + 2\\
yy' = x + 1\\
y' = \frac{x + 1}{y}\\
$$
I tried to get to the second derivative from both $yy' = x + 1$, $y' = \frac{x + 1}{y}$ and $2yy' = 2x + 2$. But every time I had that dangling constant (1 or 2), which lead to the dangling $\frac{1}{y}$ in my answer.
Like here:
$$
yy' = x + 1\\
y'y' + yy'' = 1\\
yy'' = 1 - (y')^2\\
y'' = \frac{1 - (y')^2}{y}\\
y'' = \frac{1}{y} - \frac{(y')^2}{y}\\
y'' = \frac{1}{y} - \frac{(\frac{x + 1}{y})^2}{y}\\
y'' = \frac{1}{y} - \frac{(x + 1)^2}{y^3}
$$
I don't see any way to get from my answer to the textbook's one with a transformation, no way to get rid from y in the numerator. And the correct answer doesn't have a "y" there.
Could someone either point to an error in my solution, or corroborate the suspicion that it indeed may be a typo.

From $y'y'+yy''=1$ multiply by $y^2$.
Then $(yy')^2+y^3y''=(x+1)^2+y^3y''=y^2=x^2+2x \iff y^3y''=x^2+2x-x^2-2x-1=-1$

If we continue your calculation
$y''=\dfrac 1y-\dfrac{(x+1)^2}{y^3}=\dfrac{y^2-(x+1)^2}{y^3}=\dfrac{(x^2+2x)-(x^2+2x+1)}{y^3}=\dfrac{-1}{y^3}$
Gives the same result, so I guess the textbook result is erroneous (i.e. it gives $yy''=1$ which does not agree with derivatives of $\pm\sqrt{x^2+2x}$)
