<s>[INST] Calculus Implicit Differentiation I'm learning implicit differentiation and I've hit a snag with the following equation.
$$
f(x, y) = x + xy + y = 2  
$$
$$
Dx(x) + Dx(xy) + Dx(y) = Dx(2)  
$$
$$
1 + xy' + y + y' = 0  
$$
$$
xy' + y' = -1 - y  
$$
$$
y'(x + 1) = 1 + y  
$$
$$
y' = \dfrac{(1 + y)}{(x + 1)}  
$$
$$
y'' = \dfrac{(x + 1)y' - (1 + y)}{(x + 1)^2}  
$$
$$
y'' = \dfrac{(x + 1)\dfrac{(1 + y)}{(x + 1)} - (1 + y)}{(x + 1)^2}  
$$
Ok now what according to this y'' = 0 which is wrong.
 [/INST] Your mistake seems to originate when moving from this: 
$$
xy' + y' = -1 - y  
$$
...to this, where you "lost the sign":
$$
y'(x + 1) = 1 + y$$
We need $$y'(x + 1) = -(1 + y)$$
Let's back up: $$\begin{align} 1 + xy' + y + y' & = 0 \\ \\ xy' + y' & = -1 - y\\ \\ & = -(1 + y)\end{align}$$
Then we factor out $y'$ on the left-hand side, giving us: 
$$y'(x + 1) = -(1 + y)$$
Fixing for that, then, we get: 
$$\begin{align}
y' &= \dfrac{-(1 + y)}{(x + 1)}  \\ \\
y'' & = \frac{(x + 1)(-y') - [-(1 + y)]}{(x + 1)^2} \\ \\ 
& = \frac{-(x + 1)\dfrac{-(1 + y)}{(x + 1)} + (1 + y)}{(x + 1)^2}  \\ \\
& = \frac{(1 + y) + (1+y)}{(x+1)^2}\\ \\
& = \frac{2(y+1)}{(x+1)^2}
\end{align}$$
</s>