I'm trying to solve a problem involving the implicit function theorem. I've come to a point where I must derive this implicit function with respect to it's coordinates and don't know how to proceed. Here is the problem:
Prove that exists an implicit function $\varphi$ around the point $0$ with $\varphi (0) = 0$ such that $F(\varphi(y),y)=0$
$F: \Bbb R^2 \rightarrow \Bbb R$, $F(x,y)= x^3 - x(y+1)+y^2$
Express $\varphi ' (y)$ in terms of $y$ and $\varphi(y)$
Find the Taylor polynomial of second degree of $g$ around the point $p=(1,-1)$ with $g(x.y)=\varphi(y+x^2)+xe^{y+1}$
The hypothesis for the implicit function theorem are true for $F$ around the point. And $F_x(0,0)\neq0$. So the function exists.
I have $\varphi'(y)=\frac{\varphi(y)-2y}{3 \varphi(y)^2-y-1}$
I want to find the equation for $g$'s taylor polynomial but I'm confused on how to derive with respect to what in $g$.
$g_x(x,y)=\varphi'(y+x^2)2x+e^{1+y}$ and $g_y(x,y)=\varphi'(y+x^2)+xe^{1+y}$
Could anyone help me with this? Thanks.
Update: As I have $\varphi'$ I only need to understand how to get the second order partial derivatives of $g$.
The Taylor expansion until second degree is:
$g(x,y)=g(1,-1)+(\Delta x g_x(1,-1)+\Delta yg_y(1,-1))+\frac 1 2(\Delta x^2g_{xx}(1,-1)+2\Delta x\Delta yg_{xy}(1,-1)+\Delta y^2g_{yy}(1,-1))+o\left(\lVert(\Delta x,\Delta y\rVert^3\right)$
$$\varphi'(z)(3\varphi(z)^2-z-1)=\varphi(z)-2z$$
$$\varphi''(3\varphi^2-z-1)+\varphi'(6\varphi'\varphi-1)=\varphi'-2$$
$$\varphi''=\frac{\varphi'-2-\varphi'(6\varphi'\varphi-1)}{3\varphi^2-z-1}$$
Now, $z=y+x^2=0$ for $p(1,-1)$, then $\varphi(0)=0\;;\varphi'(0)=0$ and $\varphi''(0)=2$, ready to be substituted into the expressions for the partial derivatives:
$g_x(x,y)=\varphi'(y+x^2)2x+e^{1+y}\;;g_x(1,-1)=1$
$g_{xx}(x,y)=\varphi''(y+x^2)4x^2+\varphi'(y+x^2)2\;;g_{xx}(1,-1)=8$, etc.