I'm trying to prove this using Taylor polynomials in particular, although there is a quicker way. I was hoping someone could help me out complete the proof.
Let $f : \mathbb{R}^n \rightarrow \mathbb{R}$. Suppose $f(\vec a) = 0$ and $ \frac{\partial f (\vec a)}{\partial x_n} \neq 0$ and using the implicit function theorem let $x_n$ be implicitly defined from $f= 0$ by $x_1,...,x_{n-1} = \underline x$ such that $\phi(\underline x) = x_n$. Notice therefore that $\vec x = \langle \underline x, \phi(\underline x) \rangle$ .
Prove that $\frac{\partial \phi( {\underline a)}}{\partial x_j}= -\frac{\frac{\partial f(\vec a)}{\partial x_j}}{{\frac{\partial f(\vec a)}{\partial x_n}}}$.
Notice that, using the first-order Taylor expansion for $f$ at $\vec a$, we get: $$f(\vec x) = f_{x_1}(\vec a)\Delta x_1 + \cdots + f_{x_n}(\vec a)\Delta x_n + \epsilon(\vec x)$$ Since $\frac{\partial \phi( {\underline a)}}{\partial x_j}$ amounts only to a change in $\Delta x_n$ and $\Delta x_j$, and setting the expression equal to $0$, we get: $$f(\vec x) = f_{x_j}(\vec a)\Delta x_j + f_{x_n}(\vec a)\Delta x_n + \epsilon(\vec x) = 0$$ Ignoring the error function, we get approximately: $$f_{x_j}(\vec a)\Delta x_j + f_{x_n}(\vec a)\Delta x_n + \approx 0$$ $$\frac{\Delta x_n}{\Delta x_j} \approx -\frac{f_{x_j}(\vec a)}{f_{x_n}(\vec a)}$$
Notice that $\Delta x_n$ with respect to $\Delta x_j$ is nearly the definition of $\frac{\partial \phi (\underline a)}{\partial x_j}$. What exactly would I need to get $\frac{\partial \phi( {\underline a)}}{\partial x_j}= -\frac{\frac{\partial f(\vec a)}{\partial x_j}}{{\frac{\partial f(\vec a)}{\partial x_n}}}$ from this? Especially, how could I get rid of the approximation and the error function?
This is an interesting question. To use Taylor polynomials, we assume $f$ is local $C^2$ around $\vec{a}$. Then, to calculate $\frac{\partial \phi\left(\underline{a}\right)}{\partial x_j}$, define $$ \underline{a}^{j,\Delta}=\underline{a}+\Delta e^j\,, $$ where $e^j\in\mathbb{R}^n$ and $e^j_i=\delta_{i,j}$. Thus, we need to calculate $$ \frac{\partial \phi\left(\underline{a}\right)}{\partial x_j}=\lim_{\Delta\rightarrow0}\frac{\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)}{\Delta}\,. $$
Now, we use Taylor polynomials. Because $f$ is local $C^2$ around $\vec{a}$ and $\phi$ is continuous, when $\Delta$ is small enough, $$ \begin{aligned} &\left|f((\underline{a}^{j,\Delta},\phi(\underline{a}^{j,\Delta}))-f((\underline{a},\phi(\underline{a}))-\partial_jf((\underline{a},\phi(\underline{a}))\Delta-\partial_nf((\underline{a},\phi(\underline{a}))\left(\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)\right)\right|\\ \leq& C\left(|\Delta|^2+|\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)|^2\right)\,. \end{aligned} $$ Using $f((\underline{a}^{j,\Delta},\phi(\underline{a}^{j,\Delta}))=0$, $f((\underline{a},\phi(\underline{a}))=0$, when $\Delta$ is small enough $$ \left|\partial_jf((\underline{a},\phi(\underline{a}))\Delta-\partial_nf((\underline{a},\phi(\underline{a}))\left(\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)\right)\right| \leq C\left(|\Delta|^2+|\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)|^2\right)\,, $$ which implies $$\tag{1} \left|\frac{\partial_jf((\underline{a},\phi(\underline{a}))}{\partial_nf((\underline{a},\phi(\underline{a}))}-\frac{\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)}{\Delta}\right| \leq \frac{C}{\partial_nf((\underline{a},\phi(\underline{a}))}\left(|\Delta|+\frac{|\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)|^2}{\Delta}\right)\,. $$ Thus $$ \limsup_{\Delta\rightarrow0}\left|\frac{\partial_jf((\underline{a},\phi(\underline{a}))}{\partial_nf((\underline{a},\phi(\underline{a}))}-\frac{\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)}{\Delta}\right| \leq \frac{C}{\partial_nf((\underline{a},\phi(\underline{a}))}\left(\limsup_{\Delta\rightarrow0}\frac{|\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)|^2}{\Delta}\right)\,. $$
Now, we consider two cases. First, if $$ \limsup_{\Delta\rightarrow0}\frac{|\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)|}{\Delta}<\infty\,, $$ we have $$ \limsup_{\Delta\rightarrow0}\frac{|\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)|^2}{\Delta}=0\,, $$ which implies $$ \limsup_{\Delta\rightarrow0}\left|\frac{\partial_jf((\underline{a},\phi(\underline{a}))}{\partial_nf((\underline{a},\phi(\underline{a}))}-\frac{\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)}{\Delta}\right|=0\,. $$ Second, if $$ \limsup_{\Delta\rightarrow0}\frac{|\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)|}{\Delta}=\infty\,, $$ we have a sequence ${\Delta_k}^\infty_{k=1}$ and $\lim_{k\rightarrow\infty}\Delta_k=0$ such that $$ k<\frac{|\phi\left(\underline{a}^{j,\Delta_k}\right)-\phi\left(\underline{a}\right)|}{\Delta_k}<k+1\,. $$ Go back (1) $$ \begin{aligned} &k-\left|\frac{\partial_jf((\underline{a},\phi(\underline{a}))}{\partial_nf((\underline{a},\phi(\underline{a}))}\right|\\ \leq &\left|\frac{\partial_jf((\underline{a},\phi(\underline{a}))}{\partial_nf((\underline{a},\phi(\underline{a}))}-\frac{\phi\left(\underline{a}^{j,\Delta_k}\right)-\phi\left(\underline{a}\right)}{\Delta_k}\right|\\ \leq &\frac{C}{\partial_nf((\underline{a},\phi(\underline{a}))}\left(|\Delta_k|+\frac{|\phi\left(\underline{a}^{j,\Delta_k}\right)-\phi\left(\underline{a}\right)|^2}{\Delta_k}\right)\\ \leq &\frac{C}{\partial_nf((\underline{a},\phi(\underline{a}))}\left(|\Delta_k|+|\phi\left(\underline{a}^{j,\Delta_k}\right)-\phi\left(\underline{a}\right)|(k+1)\right)\,, \end{aligned} $$ which implies for any $k>0$, $$ 1-\frac{\left|\frac{\partial_jf((\underline{a},\phi(\underline{a}))}{\partial_nf((\underline{a},\phi(\underline{a}))}\right|}{k} \leq \frac{C}{\partial_nf((\underline{a},\phi(\underline{a}))}\left(\frac{|\Delta_k|}{k}+|\phi\left(\underline{a}^{j,\Delta_k}\right)-\phi\left(\underline{a}\right)|\frac{k+1}{k}\right)\,. $$ Taking $k\rightarrow\infty$ on both sides, since $\phi$ is continuous, we obtain $$ 1\leq 0\,. $$ This is impossible, which implies the second case is impossible.