<s>[INST] 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 !
 [/INST] $$F(3+h,7+k)=4+4h+4k-5h^2-hk-7k^2+o(h^2+k^2)$$
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
$$y'(3)=a=-1\quad\text{and}\quad y''(3)=2b=\frac{5a^2+a+7a^2}2=\frac{11}2.$$
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