Let $f$ and $g$ be differentiable functions of one variable. Let $F(x,y)=f\left(x^2+g\left(x+2y\right)\right)$. Note that the relation $F(x,y)=0$ defines a function $y(x)$. Let $w = x+2y$. Show that $$\frac{\text{d}y}{\text{d}x} = -\frac{2x+g'(w)}{2g'(w)}$$ and $$\frac{\text{d}^2y}{\text{d}x^2} = \frac{2xg'(w)-2x^2g''(w)}{\left(g'(w)\right)^3}$$
I have done the first part, but in the second part I get a different answer from the desired result, namely $$\frac{\text{d}^2y}{\text{d}x^2} = \frac{\left(2g'(w)\right)\left(-2-g''(w)\left(1+2\frac{\text{d}y}{\text{d}x}\right)\right) + \left(2x+g'(w)\right)\left(2g''(w)\left(1+2\frac{\text{d}y}{\text{d}x}\right)\right)}{\left(2g'(w)\right)^2}$$ which, upon substituting in the expression for $\frac{\text{d}y}{\text{d}x}$, simplifies to $$\frac{\text{d}^2y}{\text{d}x^2} = \frac{-\left(g'(w)\right)^2 - 2x^2g''(w)}{\left(g'(w)\right)^3}$$ so please could someone check this. Thanks very much in advance for any assistance.