Consider $S \subset \mathbb{R}^3$ a regular surface. Now, since $S$ is a regular surface, a lot of theorems, propositions and definitions arise, such as
Definition. $\quad$ The quadratic form $I_p$ on $T_p(S)$, defined by $I_p(S) = |w|^2$, is called the first fundamental form of the regular surface $S \subset \mathbb{R}^3$ at $p \in S$
However, why do we need the surface to be regular? In these kind of definitions, what would go wrong if the surface is not regular?
If our patch $\bf{x}$ is regular, then we'll have that ${\bf{x}}_u\times{\bf{x}}_v$ is everywhere non-zero. Indeed, $${\bf{x}}_u\times{\bf{x}}_v=\begin{vmatrix}U_1 & U_2 & U_3\\ \partial_u x_1 & \partial_u x_2 & \partial_u x_3\\ \partial_v x_1 & \partial_v x_2 & \partial_v x_3 \end{vmatrix},$$ where $\{U_j\}$ represents the standard frame. Notice that the last two rows are the transpose of the Jacobian of $\bf{x}$ at each point. Regularity requires that this be one-to-one, so we know the above quantity is non-zero.
This is important for a few reasons. For one, it tells us that at each point $(u,v)$, the partial velocity curves are linearly independent, giving us a basis for the tangent space. From here, we can define the quantities $E={\bf{x}}_u\cdot {\bf{x}}_u,$ $F={\bf{x}}_u\cdot {\bf{x}}_v,$ and $G={\bf{x}}_v\cdot {\bf{x}}_v,$ which are the components of the first fundamental form. You'll find that $EG-F^2=\left\lVert {\bf{x}}_u\times {\bf{x}}_v\right\rVert^2,$ and requiring that this be non-zero is the same as requiring the matrix associated to the first fundamental form is invertible. We also use this to define the volume form $dS=\sqrt{EG-F^2}dudv$.
