Parameter Transformation with the Jacobian 
If  $\phi:U\rightarrow V$ and $\tilde{\phi}:\tilde{U}\rightarrow\tilde{V}$ are parametrizations of a regular surface $S$ with $V\cap\tilde{V}≠0$ and $V, \tilde{V}\subset S$. Let $E,F,G$ and $\tilde{E},\tilde{F},\tilde{G}$ be the coefficients of the I. Fundamentalform of $\phi$ and $\tilde{\phi}$ respectively. Show that;
$\left(\begin{array}{}
\tilde{E}&\tilde{F}
\\
\tilde{F}&\tilde{G}
\end{array}\right)
$=$J^t$$\left(\begin{array}{}
E&F
\\
F&G
\end{array}\right)
$$J$
in the corresponding points. $J$ is the Jacobian of $\phi^{-1}\circ\tilde\phi$

My Questions
-Can we assume that $U,\tilde U\subset\mathbb R^2$
-Is my computation for $E$ (I mean, i multiply the first column, first row of $J^t$ with E and again with the first column, first row of $J$)  below correct ?
Since, $J_{\phi^{-1}\circ\tilde\phi}(\tilde u)=J_{\phi^{-1}}(\tilde\phi(\tilde u)))\cdot J_{\tilde\phi}(\tilde u)$
In the first column, first row of $J_{\phi^{-1}}(\tilde\phi(\tilde u)))^t$ (also in the not transposed Matrix the same)  we have $\frac{1}{\phi_{u_1}(\phi^{-1}\circ\tilde\phi(\tilde u))}$ and 
In the first column, first row of $J_{\tilde\phi}(\tilde u)^t$we have $\tilde\phi_{\tilde u_1}(\tilde u)$
Finally;
$(\frac{1}{\phi_{u_1}(\phi^{-1}\circ\tilde\phi(\tilde u))})^2(\tilde\phi_{\tilde u_1}(\tilde u))^2\langle\phi_{u_1},\phi_{u_1}\rangle=\langle\tilde\phi_{\tilde u_1},\tilde\phi_{\tilde u_1}\rangle$
 A: *

*Yes, you can assume that. A parameterization of a surface is (generally) a map from an open set in the plane to the surface.

*I confess, in the remaining computation, I'm having a hard time seeing what you're doing. That may be because of the notation -- all those tildes and inverses are tough on my old eyes. It looks as if you've applied the "derivative of the inverse" rule at the end of the first line saying "In the first column". If that's what's going on, then it appear that what you've written is correct. 
I'd like to suggest a notationally simpler alternative, assuming for the moment that for you, a "regular surface" is a surface embedded nicely in $R^n$.
First, note that the $ij$ entry of the matrix $AB$ is the inner product of the $i$th row of $A$ with the $j$th column of $B$. So the $ij$ entry of $A^t A$ is the inner product of the $i$th and $j$th columns of $A$.  
If $D(u)$ is the derivative of the function $\phi$ at the point $u \in U$, i.e., it's an $n \times 2$ matrix, then your first fundamental form (for the first parameterization) is just $D(u)^t D(u)$, by the preceding paragraph. Call this "fact 1." A corresponding statement holds for the "tilde" variables. 
Letting $\psi = \phi^{-1}\circ\tilde\phi$, so that $\phi \circ \psi = \tilde \phi$ the chainrule tells us that 
$$
\tilde D(u') = D(u) \cdot J_\psi(u)
$$
where $u' = \psi(u)$, and $J$ is the Jacobian. Writing "FF" for the fundamental form, we have
$$
\tilde{FF}(u') = \tilde D(u')^t \tilde D(u')
$$ 
by fact 1. Doing a substitution, we get
\begin{align}
\tilde{FF}(u') 
&= \tilde D(u')^t \tilde D(u')\\
&= \left(D(u) J_\psi(u)\right)^t \left(D(u) J_\psi(u)\right)\text{, by fact 1}\\
&= \left(J_\psi(u)^t D(u)^t\right)  \left(D(u) J_\psi(u)\right)\\
&= J_\psi(u)^t \left(D(u)^t  D(u)\right) J_\psi(u)\\
&= J_\psi(u)^t FF(u) J_\psi(u)
\end{align}
which is the conclusion you wanted. And all I used was matrix multiply, the chain rule, and some algebra. 
One reason to favor this approach is that it gets used over and over again, so it's nice to get used to it early. 
