Say what it means for a system of equations
$$f_1 = \cdots = f_m = 0,$$
where $f_i(x_1, \cdots x_n)$ are differentiable functions, to define a submanifold near a point $a = (a_1, \cdots a_n), f_i(a) = 0, i = 1, \cdots m$.
From my understanding, a submanifold is patches of various surfaces put together, but I don't see how I can get that definition from this.
I'm not sure if this is what you mean, but let $\operatorname{f} : \mathbb{R}^{n} \to \mathbb{R}^m$, where $n > m$. The $m$-by-$n$ Jacobian matrix is given by:
$$\left[ \begin{array}{ccc} \partial\operatorname{f}_1/\partial x_1 & \cdots & \partial\operatorname{f}_1/\partial x_n \\ \vdots & \ddots & \vdots \\ \partial\operatorname{f}_m/\partial x_1 & \cdots & \partial\operatorname{f}_m/\partial x_n \end{array}\right]$$
A point $x \in \mathbb{R}^n$ is called a regular point if the Jacobian matrix has maximal rank, i.e. rank $m$, when evaluated at $x$. A point $y \in \mathbb{R}^m$ is called a regular value if all of the points $x \in \mathbb{R}^n$ for which $\operatorname{f}(x)=y$ are regular points. In such a case, the pre-image of a regular value is a smooth parametrisable manifold.
In your case, if all of the $x \in \mathbb{R}^n$ for which $\operatorname{f}(x)=0$ are regular points then the level-set $\operatorname{f}^{-1}(0)$ is a smooth parametrisable manifold. This is the Implicit Function Theorem.
In practise: Find all of the $x \in \mathbb{R}^n$ (critical points) for which the Jacobian has rank less than $m$. Then find all of the images $\operatorname{f}(x)$ (critical values). Just check that $0 \in \mathbb{R}^m$ is not a critical value.