Find derivative using implicit function theorem 
Let $F(x,y)$ be a function such that
$$F(x,y)=4+4(x -3) +4(y-7)-5(x-3)^2-(x-3)(y-7)-7(y-7)^2 +R_2$$
is the Taylor series.
Using implicit function theorem and find $y'(3),y''(3)$ of the equation $F(x,y)=4$ in $(3,7)$.

My attempt :
Using implicit function theorem , I know that
$$y'(3)=-\frac{f_x}{f_y}=\left(-\frac{4-10\left(x-3\right)-y+7}{4-x+3-14\left(y-7\right)}\right)=-1.$$
Then
$$y''(3)=-\frac{139y-1009}{\left(-x+105-14y\right)^2}=-\frac{139\cdot 7-1009}{\left(-3+105-14\cdot 7\right)^2}=\frac{9}{4}.$$
$y'(3)$ is correct but $y''(3)$ is wrong but I don't get why.
Thanks !
 A: Let $F=f+R_2$, then, in your work, taking the derivative with respect of $x$ of $-f_x/f_y$, you should consider $y$ as a function of $x$:
$$\frac{d}{dx}\left(-\frac{4-10\left(x-3\right)-(y-7)}{4-(x-3)-14\left(y-7\right)}\right)\\
=-\frac{(-10-y')(4-(x-3)-14\left(y-7\right))-(4-10\left(x-3\right)-(y-7))(-1-14y')}{(4-(x-3)-14\left(y-7\right))^2}.$$
Letting $x=3$, $y=7$ and $y'=-1$, we find
$$y''(3)=-\frac{(-10-(-1))(4)-(4)(-1-14(-1))}{(4)^2}=\frac{11}{2}.
$$
P.S. We should also note that $y'(3)$ and $y''(3)$ DO NOT depend on the remainder $R_2$. Indeed, we have that
$$y'(x)=-\frac{F_x}{F_y}$$
and by the chain rule
$$y''(x) =\frac{\partial y'}{\partial x}\left(-\frac{F_x}{F_y}\right)  + \frac{\partial y'}{\partial y}\left(-\frac{F_x}{F_y}\right) y'(x) = \frac{-F_y^2F_{xx}+ 2F_xF_yF_{xy}-F_x^2F_{yy}}{F_y^3}.$$
In our case, from the expansion of $F$ we obtain that at $(3,7)$,
$$F_x=F_y=4,\; F_{xx}=-10,\;F_{xy}=-1,F_{yy}=-14,$$
and therefore
$$y'(3)=-\frac{4}{4}=-1\quad y''(3) = \frac{-4^2(-10)+ 2\cdot 4 \cdot 4 (-1)-4^2 (-14)}{4^3}=\frac{11}{2}.$$
A: $$F(3+h,7+k)=4+4h+4k-5h^2-hk-7k^2+o(h^2+k^2)$$
and
$$y(3+h)=7+ah+bh^2+o(h^2),\quad\text{where}\quad a=y'(3)\quad\text{and}\quad b=\frac{y''(3)}2.$$
Since $F(3+h,y(3+h))=4,$ we deduce:
$$0=4h+4(ah+bh^2)-5(ah)^2-h(ah)-7(ah)^2,$$
hence
$$y'(3)=a=-1\quad\text{and}\quad y''(3)=2b=\frac{5a^2+a+7a^2}2=\frac{11}2.$$
