Two equations implicitly determine two variables near the origin

Given: Put $F_1=x^2+y^2-(z+1{)}^3+1$ and $F_2=e^x+e^y+e^z-3$. Sums and scalar multiples use Sums and scalar multiples of totally differentiable maps are totally differentiable with the expected derivatives and Sums, scalar multiples, products and quotients: $(f+g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$, $(\alpha f{)}^{\mathrm{\prime }}(c)=\alpha f^{\mathrm{\prime }}(c)$, $(fg{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)g(c)+f(c)g^{\mathrm{\prime }}(c)$, and $(f\mathrm{/}g{)}^{\mathrm{\prime }}(c)=(f^{\mathrm{\prime }}(c)g(c)-f(c)g^{\mathrm{\prime }}(c))\mathrm{/}g(c{)}^2$ when $g(c)\mathrm{\ne }0$.

Let $k,m,n\ge 1$ and $p\in \mathbb{N}$, let $U\subseteq {\mathbb{R}}^{m+n+p}$ be open, and let $F:U\to {\mathbb{R}}^n$ be $C^k$. If $F(a,b,{\lambda }_0)=0$ and $D_{y}F(a,b,{\lambda }_0)$ is invertible, then on suitable neighbourhoods there is a unique $C^k$ map $\phi$ with $F(x,y,\lambda )=0$ exactly when $y=\phi (x,\lambda )$, and $D\phi =-D_y{F}^{-1}{D}_{(x,\lambda )}F$ along its graph (The parametrized implicit function theorem with $C^k$ regularity).

The real exponential function is $C^{\mathrm{\infty }}$ and has derivative equal to itself (The exponential function is smooth and $(\exp {)}^{\mathrm{\prime }}=\exp$).

For every integer $d\ge 1$, the derivative of $x^d$ is $d{x}^{d-1}$ (For a natural $n\ge 1$ the function $x\mapsto x^n$ is differentiable everywhere with derivative $\iota (n)x^{n-1}$; for $n=0$ it is the constant $1$, with derivative $0$; for a natural $n\ge 1$ the function $x\mapsto x^{-n}$ is differentiable at every $x\mathrm{\ne }0$ with derivative $-\iota (n)x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term).

The derivative of a composite is the product of the outer derivative at the inner value and the inner derivative (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$, then $f\circ g$ is differentiable at $c$ with $(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)$).

Finite componentwise sums, products, scalar multiples, and composites of $C^k$ Euclidean maps are $C^k$ ($C^k$ Euclidean maps are closed under componentwise algebra and composition).