I have $(f_1, f_2) : \mathbb{R}^n \to \mathbb{R}^2$ and say at $0$ the Hessian say at $(\mathbf{a}, \mathbf{b}) = (a_1, .., a_{n-2}, b_1, b_2)$ we have $(f_1(\mathbf{a}, \mathbf{b}), f_2(\mathbf{a}, \mathbf{b})) = (0,0)$ and $$ \begin{pmatrix} \partial f_1/\partial y_1 (\mathbf{a}, \mathbf{b}) & \partial f_1/\partial y_2 (\mathbf{a}, \mathbf{b}) \\ \partial f_2/\partial y_1 (\mathbf{a}, \mathbf{b}) & \partial f_2/\partial y_2 (\mathbf{a}, \mathbf{b}) \end{pmatrix} $$ is invertible where we denote the $n$ variables by $(x_1, .., x_{n-2}, y_1, y_2)$. Then we can apply the implicit function theorem to obtain $g:U \to \mathbb{R}^2$, where $U$ is an open neighborhood of $\mathbf{a}$, such that $(f_1(\mathbf{x}, g(\mathbf{x}), f_2(\mathbf{x}, g(\mathbf{x})) = (0, 0)$ on $U$.
I want to find a higher partial derivatives of $g$. If there was only one function $f_1$ then I could take a derivative of $f_1$, set it to $0$ and get an expression for $\partial f_1/ \partial x_i$ and then take a derivative again. What can I do when there is more than one function as in this case? Thank you.
The expression of the second differential is in general tedious. I write it in a general setting below.
Let $E,F$ be Banach spaces, $\mathcal{U}\subset E$ an open set and $f\in\mathcal{C}^k(\mathcal{U},F)$ for some $k\geq 2$. We assume that $E=E'\times E'$ and that the partial differential $\mathrm{d}_2f(x_0):=\mathrm{d}f(x_0)_{\vert E''}\in\mathcal{L}(E'',F)$ is invertible for some $u_0\in\mathcal{U}$. By the implicit function theorem, there is a neighborhood $\mathcal{V}'\times\mathcal{V}''\subset\mathcal{U}$ of $x_0$ such that $f(x',x'')=\beta:=f(x_0)$ if and only if $x''=g(x')$ for some $g\in\mathcal{C}^k(\mathcal{V}'',F)$. Differentiating this equation for a fixed $\beta\in F$ yields \begin{align*} \mathrm{d}_1f(x',g(x'))+\mathrm{d}_2f(x',g(x'))\circ\mathrm{d}g(x')&=0 \end{align*} and thus \begin{align*} \mathrm{d}g(x')&=-\mathrm{d}_2f(x',g(x'))^{-1}\circ\mathrm{d}_1f(x',g(x'))\in\mathcal{L}(E',E''). \end{align*}
We now look for an expression of $\mathrm{d}^2g(x')\in\mathcal{L}(E'\times E',E'')$ (which makes sense as $k\geq 2$). We let $h\in E'$ and \begin{align*} L&:E'\ni x\longmapsto L(x'):=\mathrm{d}_2f(x',g(x'))^{-1}\in\mathcal{L}(F,E''),\\ % M&:E'\ni x\longmapsto M(x'):=\mathrm{d}_1f(x',g(x'))[h]\in F \end{align*} so that \begin{align*} \mathrm{d}L(x')&\in\mathcal{L}(E',\mathcal{L}(F,E''))\simeq\mathrm{Bil\,}(E'\times F,F),\\ % \mathrm{d}M(x')&\in\mathcal{L}(E',F). \end{align*} For all $k\in E'$, we have \begin{align*} \mathrm{d}\big(L(x')\circ M(x')\big)[k]&=(\mathrm{d}L(x')[k])\circ M(x')+L(x')\circ(\mathrm{d}M(x')[k])\in E''. \end{align*} Now we compute for all $(h,\ell)\in E'\times F$: \begin{align*} \mathrm{d}L(x')&=-\mathrm{d}_2f(x',g(x'))^{-1}\circ\bigg[\mathrm{d}_1\big(\mathrm{d}_2f(x',g(x'))\big)+\mathrm{d}_2\big(\mathrm{d}_2f(x',g(x'))\big)\circ\mathrm{d}g(x')\bigg]\circ\mathrm{d}_2f(x',g(x'))^{-1},\\ % \mathrm{d}L(x')[h,\ell]&=-\mathrm{d}_2f(x',g(x'))^{-1}\bigg[\mathrm{d}_1\big(\mathrm{d}_2f(x',g(x'))\big)\Big[\mathrm{d}_2f(x',g(x'))^{-1}[h],\ell\Big]\\ &\qquad\qquad\qquad\qquad\quad+\mathrm{d}_2\big(\mathrm{d}_2f(x',g(x'))\big)\Big[\mathrm{d}_2f(x',g(x'))^{-1}[h],\mathrm{d}g(x')[\ell]\Big]\bigg],\\ % % \mathrm{d}M(x')[k]&=\mathrm{d}_1\big(\mathrm{d}_1f(x',g(x'))\big)[h,k]+\mathrm{d}_2\big(\mathrm{d}_1f(x',g(x'))\big)\Big[h,\mathrm{d}g(x')[k]\Big]. \end{align*} We finally obtain for all $(h,k)\in E'\times E'$ (replacing $\ell$ by $\mathrm{d}_1f(x',g(x'))[h]\in F$): \begin{align*} \mathrm{d}^2g(x')[h,k]&=\mathrm{d}_2f(x',g(x'))^{-1}\bigg[\mathrm{d}_1\big(\mathrm{d}_2f(x',g(x'))\big)\Big[\mathrm{d}_2f(x',g(x'))^{-1}[h],\mathrm{d}_1f(x',g(x'))[h]\Big]\\ &\qquad\qquad\qquad\qquad\quad+\mathrm{d}_2\big(\mathrm{d}_2f(x',g(x'))\big)\Big[\mathrm{d}_2f(x',g(x'))^{-1}[h],\mathrm{d}g(x')\big[\mathrm{d}_1f(x',g(x'))[h]\big]\Big]\bigg]\\ % &-\mathrm{d}_2f(x',g(x'))^{-1}\bigg[\mathrm{d}_1\big(\mathrm{d}_1f(x',g(x'))\big)[h,k]+\mathrm{d}_2\big(\mathrm{d}_1f(x',g(x'))\big)\Big[h,\mathrm{d}g(x')[k]\Big]\bigg]\in E''. \end{align*}
In your case, you have to replace differentials by partial derivatives and matrices.