Long, complicated implicit differentiation, finding $y'$ Find $y''$ where $y^2 + xy = \ln (x+1)$
Using implicit differentiation,
$\frac{dy}{dx} = \frac{1}{2y+x} (\frac{1}{x+1} -y)$
Now finding the second derivative of this function, using the product rule,
$\frac{d^2y}{dx^2} = (\frac{-2 \frac{dy}{dx} +1}{(2y+x)^2}  \times \frac{1}{x+1} - y) + (\frac{1}{2y+x} \times (\frac{-1}{(x+1)^2} - \frac{dy}{dx}))$
This looks very different from what the answer key is telling me:
$\frac{d^2y}{dx^2} = \frac{-1}{2y+x}(\frac{1}{(1+x)^2} + 2\frac{dy}{dx} + 2(\frac{dy}{dx})^2)$

Another method: consider your equation
$$y^2 + xy - \ln (x+1)=0$$
as a quadratic in $x$ giving:
$$y=\frac12 \left(-x \pm \sqrt{x^2+4 \ln(x+1)}\right)\tag{1}$$
Then differentiate twice.
In an alternate way, write (1) under the equivalent form:
$$(2y+1)^2=(x^2+4 \ln(x+1))^2$$
then differentiate twice,
giving simpler calculations.
