How do you show that some given equations define a k-dimensional submanifold in an open neighborhood of the origin?
For example, I am given the equations
$f_1(\mathbf{x}) = e^{x_1} + e^{x_2} + e^{x_3} + x_1x_2x_3 - 3,$
$f_2(\mathbf{x}) = x_1 + x_2^2 + x_3^3 + \sin\left(e^{x_1x_2x_3}-1\right).$
and I want to show that $f_1(\mathbf{x}) = f_2(\mathbf{x}) = 0$ defines a one-dimensional submanifold in an open neighborhood of the origin.
What is the general approach for this kind of problem?
The general approach is that there's one big theorem: the Implicit Function Theorem. You should be familiar with the special case of a function from $\mathbb{R}^2$ to $\mathbb{R}^1$ from a basic calculus class; this generalizes that.
We have a differentiable function $f$ from some open subset of $\mathbb{R}^n$ to $\mathbb{R}^m$, such that $f(a)=b$. Look at the $m\times n$ derivative matrix $f'(a)$. If the columns of $f'(a)$ span $\mathbb{R}^m$, then the level set $\{x:f(x)=b\}$ is locally a $(n-m)$-dimensional manifold near $a$. Also, if we choose $m$ linearly independent columns from $f'(a)$, we can locally parametrize the level set in terms of the variables corresponding to the other columns.
So, your example. The function $f\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} = \begin{pmatrix}e^{x_1}+e^{x_2}+e^{x_3}+x_1x_2x_3-3\\ x_1+x_2^2+x_3^3+\sin(e^{x_1x_2x_3}-1)\end{pmatrix}$ has derivative $$f'\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}e^{x_1}+x_2x_3&e^{x_2}+x_1x_3&e^{x_3}+x_1x_2\\ 1+x_2x_3e^{p}\cos(e^{p}-1)&2x_2+x_1x_3e^{p}\cos(e^{p}-1)&3x_3^2+x_1x_2e^{p}\cos(e^{p}-1)\end{pmatrix}$$ (abbreviating $x_1x_2x_3$ as $p$). Evaluate at the origin, and $$f'\begin{pmatrix}0\\0\\0\end{pmatrix}=\begin{pmatrix}1&1&1\\ 1&0&0\end{pmatrix}$$ The first two columns are linearly independent, so we can write $x_1$ and $x_2$ as functions of $x_3$ in the level set here. We can't explicitly write down formulas for those functions, but we can differentiate them implicitly.