showing that two surface is isometric I tried to show that a parametrized surface $S$ in $\mathbb{R}^3$ given by $(u, v)$ ->$(u, v, u^2)$ is isometric to the flat plane.
At first, I found their first fundamental form, but they are different.
(I choose a surface patch defined by $(u, v)$ -> u$\mathbb{q}$ + v$\mathbb{p}$ where $\mathbb{p}$ and $\mathbb{q}$ are perpendicular and both unit vectors.)
In presseley's text, there is a theorem following that two surface are local isometric if and only if they have the same first fundamental form.
So I concluded that they are not isometric
How can I show that?
 A: Let $E$ be the plane defined by $z=0$. In your case you may consider $f\colon R\to R$ defined by 
$$f(x)=-\frac{1}{4}\left(\ln(-2x+\sqrt{1+4x^2})-2x\sqrt{1+4x^2}\right).$$
Verify that $f'(x)=\sqrt{1+4x^2}$.  Define $\phi\colon S\to E$, $\phi(u,v,u^2):=\bigl(f(u),v,0\bigr)$.  It's easy to verify that $\phi$ is an isometry; check the FFFs.
I've constructed $f$ in an obvious way.
Edit (much later): The construction of $f$ is obvious since the FFF of the given map is
$$\begin{pmatrix}1+4u^2& 0\\ 0&1\end{pmatrix}.$$
A: I think that proof can be found on various textbook. The following one is one : 
$$ {\bf x} : U \rightarrow S,\ \overline{{\bf x}} : U\rightarrow \overline{S} $$ 
Consider $\phi = \overline{{\bf x}}\circ {\bf x}^{-1}$. That is, $$
  d\phi (a{\bf x}_u +b{\bf x}_v)\cdot d\phi (a{\bf x}_u + b{\bf x}_v) = (a\overline{{\bf x}}_u + b\overline{{\bf x}}_v)\cdot (a\overline{{\bf x}}_u + b\overline{{\bf x}}_v) = a^2 \overline{E} + 2ab \overline{F} + b^2\overline{G}.$$
So $d\phi$ is local isometric $\Leftrightarrow a^2 \overline{E} + 2ab \overline{F} + b^2\overline{G} =a^2E+ 2ab F + b^2G \ (\forall\ a,\ b) \Leftrightarrow \overline{E}=F,\  \overline{F}=F,\ \overline{G} =G$. 
