finding derivatives of variables in multivariable taylor polynomial Given:
$$F(x,y) = -6 -4(x-4) +6(y-6) +8(x-4)^2 +9(x-4)(y-6) -4(y-6)^2 + R_2$$
and that $F(4,6)=-6$ 
find y'(4), y''(4), make sure that the function is applicable for the implicit function theorem.
 my attempt:
so I've basically made sure that the conditions applied, implicitly derived F and found out y'. It was a lot of hard work and while I got a result my method didn't at all utilize the fact it's taylor -I think there must be an easier way but I just can't put my finger on it. I think it'll utilize the fact that $-4 = F_x(x,y)$ and that  $6 = F_y(x,y)$ - and maybe the chain rule?
since $$-4 = \frac{\partial F(4,6)}{\partial{x}}=\frac{\partial{y}}{\partial{x}}* \frac{\partial F(4,6)}{\partial{y}} = 6*y'(x) \Rightarrow y'(x) = \frac{-2}3$$
but both my calculation and this calculator https://www.emathhelp.net/en/calculators/calculus-1/implicit-differentiation-calculator/?f=-4%28x-4%29%2B6%28y-6%29%2B8%28x-4%29%5E2+%2B9%28x-4%29%28y-6%29+-4%28y-6%29%5E2%3D0&type=x&px=4&py=6 output $2/3$ 
To sum up - where is my logic faulty? is there a similar easier way? how do I extend this to y''(4)?
 A: From the fact that
$$
F(x,y)=F(4,6)+F_x(4,6)(x-4)+F_y(4,6)(y-6)+\frac12F_{xx}(4,6)(x-4)^2
$$
$$
+\frac12F_{yy}(4,6)(y-6)^2+F_{xy}(4,6)(x-4)(y-6)+R_2
$$
it follows that
$$
F_x(4,6)=-4,\; F_y(4,6)=6,\; \frac12F_{xx}(4,6)=8,\; F_{xy}(4,6)=9,\; 
\frac12 F_{yy}(4,6)=-4. 
$$
Then the derivative of the implicit function is
$$
\left.y'(x)\right|_{(4,6)}=-\frac{F_x(4,6)}{F_y(4,6)}=-\frac{-4}{6}=\frac23.
$$
Update begin
This formula for the derivative of an implicit function of two variables follows from the multivariable chain rule. $F(x,y(x))\equiv C$ implies that
$$
(F(x,y(x)))'=0
$$
or
$$
\frac{\partial F}{\partial x}\cdot 1+\frac{\partial F}{\partial y}\cdot y'(x)=0;
$$
therefore,
$$
y'(x)=-\left. \frac{\partial F}{\partial x} \right/ \frac{\partial F}{\partial y}.
$$
Update end
To calculate the second-order derivative, we can use the chain and quotient rules:
$$
y''(x)=(y'(x))'=\left(\left.\left( -\frac{F_x}{F_y} \right)\right|_{y=y(x)}\right)'= 
\left( -\frac{F_x(x,y(x))}{F_y(x,y(x))} \right)'
$$
(mostly skip $(x,y(x))$ for brevity)
$$
(F_x(x,y(x)))'=F_{xx}\cdot 1+F_{xy}\cdot y'(x)
=F_{xx}-F_{xy}\cdot\frac{F_x}{F_y}
$$
$$
(F_y(x,y(x)))'=F_{yx}\cdot 1+F_{yy}\cdot y'(x)
=F_{yx}-F_{yy}\cdot\frac{F_x}{F_y}
$$
Thus,
$$
y''(x)=-\frac{(F_x(x,y(x)))'F_y-F_x (F_y(x,y(x)))'}{(F_y)^2}
$$
$$
=-\frac{\left(F_{xx}-F_{xy}\cdot\frac{F_x}{F_y}\right)F_y-F_x \left(F_{yx}-F_{yy}\cdot\frac{F_x}{F_y}\right)}{(F_y)^2}.
$$
So
$$
\left.y''(x)\right|_{(4,6)}=-\frac{\left(16+9\cdot\frac23\right)6+4 \left(9-8\cdot\frac23\right)}{36}.
$$
