Implicit Function Theorem in Higher Dimensions The following function is given: $$f:\Bbb R^3 \rightarrow \Bbb R^2, \left(\begin{matrix}x \\ y \\ z\\ \end{matrix}\right) \mapsto \left(\begin{matrix}-2x^2 + y^2 + z^2 \\ x^2+e^{y-1} - 2y \end{matrix}\right)$$
The first task is to determine whether the function can be solved in terms of $y$ and $z$ at the point $(1, 1, 1)^T$, when $f = (0, 0)^T$. Should this be possible, the next task would be to calculate the derivative at the aforementioned point.
Attempt at a solution:
$f\left(\begin{matrix}1 \\ 1 \\ 1\\ \end{matrix}\right) = \left(\begin{matrix} 0 \\ 0\\ \end{matrix}\right)$ is clear.
Calculate the determinant of the following Jacobian matrix $$\left| \begin{matrix} \frac{\partial f_1(1,1,1)}{\partial y} && \frac{\partial f_1(1,1,1)}{\partial z} \\ \frac{\partial f_2(1,1,1)}{\partial y} && \frac{\partial f_2(1,1,1)}{\partial z}\end{matrix}\right|,$$where $f_1(x,y,z)=-2x^2 + y^2 + z^2$, $f_2(x,y,z)=x^2+e^{y-1} - 2y$.
Thus, we have$$\left| \begin{matrix} 2y^2 && 2z^2 \\ e^{y-1}-2 && 0 \end{matrix}\right|_{(1,1,1)}=\left| \begin{matrix} 2 && 2 \\ -1 && 0 \end{matrix}\right|=2 \neq 0$$
Thus, by the implicit function theorem, there exist open neighborhoods $U\subseteq \Bbb R$ and $V\subseteq \Bbb R^2$ with $1\in U$ and $(1,1)^T \in V$ and a continuously differentiable function $g:U \rightarrow V$ such that for all $(x,y) \in U \times V$ the following holds:$$f(x,y) = 0 \iff y=g(x)$$
It's the next task that I'm not 100% sure on. Do I need to calculate the partial of $f$ w.r.t. $y$ and then again w.r.t. $z$? How does one do this?
 A: Suppose you know that $x = 9/10$ (a representative number near $x = 1$, i.e.,  a candidate for an element of $U$). Then if $f(x, y, z) = 0$, what do you know about $y$ and $z$? From the second term, you know that 
$$
(9/10)^2+e^{y-1} - 2y = 0\\
0.81 + e^{y-1} - 2y = 0 
$$
If you're willing to guess that the solution for $y$ is near $1$, you can approximate $e^{y-1}$ with $1 + (y-1)$ (the first two terms of the Taylor series) to convert this to 
$$
0.81 + (1 +  (y-1))  - 2y \approx 0 \\
0.81 \approx y
$$
thus determining $y$ from a known value of $x$. 
More generally, you can see that there's a unique solution for $y$: the one-dimensional implicit value theorem applied to $f_2(x, y)$ near the point $(x, y) = (1, 1)$ says so, since $\frac{\partial f_2}{\partial y} (1, 1) = -1$, as you already computed. So there's a function $h$, defined on a neighborhood $U$ of $x = 1$, with the property that 
$$
f_2(x, h(x)) = 0
$$
for $x \in U$. 
Now continuing with the example,  knowing $x = 9/10$ and $y \approx 0.81 $, look at the first term: from that, you can solve for $z$. It'll be a square root of some kind, and one of the two roots will be near $+1$ and the other near $-1$ so you pick the $+1$ root. 
Continuing with the general analysis instead of the single instance, we have that
$h(x)$ is a number such that $x^2 + e^{h(x) - 1} - 2h(x) = 0$ (for $x$ near $0$); we can then build the required function $g$ via
$$
g(x) = \begin{bmatrix} h(x) \\ \sqrt{2x^2 - h(x)^2}  \end{bmatrix}
$$
Does that help? The fact that you can't explicitly write $h$ isn't a problem -- you know from the 1D implicit function theorem that it exists. 
A: I think what you want is a function $(y,z)=g(x)$, that is a curve in a parametric representation instead of the intersection of two surfaces $f_1(x,y,z)=0, f_2(x,y,z)=0$.
In your statement: 
" ...$g:U→V$ such that for all $\underline{(x,y)\in U\times V}$ the following holds: $\underline{f(x,y)=0}$ iff  $\underline{y=g(x)}$..."
you should probably say: $(x,(y,z))\in U\times V$ and $f(x,y,z)=(0,0)$ iff $(y,z)=g(x)$.
Your calculation of the condition of existence is correct.
This being cleared out you just do brute force calculation, pretending $f(x,g(x))=0$ wherever $g(x)$ is defined. Then
 $0=\frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial (y,z)}\cdot \frac{dg}{dx}$ and you can solve it for the vector $\frac{dg}{dx}$
plugging in the values of $x,y,z$ in your point.
