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?
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.