Question: (i) Prove that the system of equations $$F_1(x, y, z) := 3x + e y - cos z + z = 0$$ and $$F_2(x, y, z) := sin x - x - y + cos y + z - 1 = 0$$ defines implicitly in a neighborhood of the origin, a curve with parametric equations $x = f_1(y)$ and $z = f_2(y)$.
(ii) Write down the parametric equations of the tangent line to the curve in (i) at the point $(0, 0, 0)$.
I've attempted to answer (i) but I'm new to the implicit function theorem and I'm not sure if I've completely miss used it. Also if somebody could show me how to answer (ii) that would be really appreciated.
Attempted Answer: Clearly the implicit function theorem needs to be used. So to start we note that $F_1(0,0,0)=0$ and $F_2(0,0,0)=0$. Next note that the Jacobian \begin{pmatrix} \frac{\partial F_1}{\partial x}&\frac{\partial F_1}{\partial y} \\ \frac{\partial F_2}{\partial x} & \frac{\partial F_2}{\partial y} \end{pmatrix}equals
\begin{pmatrix} 3&e^y \\ \cos x -1& -1 -\sin x \end{pmatrix} which evaluated at $(0,0,0)$ has a non-zero determinant so by the implicit function theorem there exists $\alpha, \beta >0$ and two unique functions $f_1:B(y,\alpha) \rightarrow B(x,\beta)$ and $f_1:B(y,\alpha) \rightarrow B(z,\beta)$ such that $F_1(f_1(y'),y',f_2(y'))=0$ and $F_2(f_1(y'),y',f_2(y'))=0$ for all $y' \in B(y,\alpha)$.