Finding the second derivative; What am I doing wrong? Original Question:
$xy+y-x=1$
Find the second derivative;
$d^2y\over{dx^2}$$(xy+y-x=1)$
We are allowed to use either notation as far as I know: ${dy\over{dx}}$ or ${y'}$. Because ${dy\over{dx}}=y'$ according to my Math 1000 prof.
$(xy)'+y'-x'-1'=0'$
$(y+xy')+y'-1=0$
$y'(x+1)-1+y=0$
$y'(x+1)=1-y$
$y'={1-y\over{x+1}}$
Now we're supposed to find $y''$ but, that's where I mess things up. Quotient rule:
$y''={{(1-y)'(x+1)-(1-y)(x+1)'}\over[(x+1)]^2}$
$y''={{(-y')(x+1)-(1-y)(1)}\over[(x+1)]^2}$
$y''={{-({{1-y}\over{x+1}}){\cdot}(x+1)-(1-y)}\over{(x+1)^2}}$
$y''={{-(1-y)-(1-y)}\over{(x+1)^2}}={-2(1-y)\over{(x+1)^2}}={2(y-1)\over{(x+1)^2}}$
But, The answer is supposed to be ${y''={{y-1}\over{(x+1)^2}}}$
Why am I getting a 2? That's all I want to know.
Also, I took out the unnecessary equal signs, my prof wants us to have those cause he's basically insane but, it's confusing everyone so I'm taking them out.
The improper method! According to my prof:
$(xy)'+y'-x'-1'=0'$
$(y+xy')+y'-1=0$
$y'+(xy')'+y''=0$
$y'+(y'+xy'')+y''=0$
$y''(x+1)=-2y'$
${y''={{-2y'}\over{x+1}}}$
Plugin $y'$ and it's the same answer i'm getting. Supposedly the wrong one everywhere I've looked.

Answer:
I will tell you why you have been confused.  I looked at your Wolfram Answer and you have wrongly specified such as
derivative of $\frac{(1-y)}{(1+x)}$ .  Wolfram Alpha thinks that it is the partial derivative of this and treats y as a constant and merely calculates $(1-y).\frac{d}{dx}\frac{1}{(1+x)}$ and this ofcourse will be
${(1-y)}\frac{-1}{(1+x)^{2}}$ and eventually spits out $\frac{(y-1)}{(1+x)^{2}}$ and you think you have been wrong.
Now specify in wolfrom alpha the following way:
$\frac{d}{dx} ((1+x)\frac{dy}{dx}+(y-1))$  This is valid because y'(1+x) + y-1 =0, Now you have to find y''
and see what solution you get.  You will be right on track.
https://www.wolframalpha.com/input/?i=d%2Fdx+%28%281%2Bx%29dy%2Fdx%2B%28y-1%29%29
Good luck
