If $f:U\rightarrow R$ is differentiable, $U\subset R^2$, and graph(f) is a regular surface, why is? 
If $f:U\rightarrow \mathbb{R}$ is differentiable, $U\subset \mathbb{R}^2$, and $\operatorname{graph}(f)=:S$ is a regular surface,  why is ?
$1+||\nabla  f||^2=||f_x||^2||f_y||^2-2\langle f_x,f_y\rangle^2$

The original Question was to prove that;
$\operatorname{Area}(S)=\int_U\sqrt{1+|| \nabla f||^2}$
but in the lecture, we've seen that the area is $\int_U \det(g_{ij})^{1/2}$, where $(g_{ij})$ is the First Fundamental Form, So this is exactly the RHS of the equation.
Edit: Thank you very very much Fantini.
 A: Graphs of functions have a parametrization given by
$$\mathbf{r}(x,y) = (x,y,f(x,y)).$$
The area of a surface is defined as
$$\operatorname{Area}(S) = \iint\limits_{R} \left\vert \frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial y} \right\vert \, dA,$$
where $R$ is the projected region in the plane. Once you compute that you will find
$$\frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial y} = ( - f_x, - f_y, 1 ),$$
therefore
$$\left\vert \frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial y} \right\vert^2 = 1 + (f_x)^2 + (f_y)^2 = 1 + ||\nabla f||^2.$$
This is the usual calculus way of doing it. Using differential geometry you need to compute $\sqrt{EG - F^2}$, where $E, G$ and $F$ are defined by
$$
\begin{align}
E & = \langle \mathbf{r}_x, \mathbf{r}_x \rangle, \\
G & = \langle \mathbf{r}_y, \mathbf{r}_y \rangle, \\
F & = \langle \mathbf{r}_x, \mathbf{r}_y \rangle.
\end{align}
$$
Notice that this is not equivalent to what you have written, because $\mathbf{r}_x \neq f_x$ and likewise for $y$. We have
$$\mathbf{r}_x = (1,0,f_x) \text{ and } \mathbf{r}_y = (0,1,f_y),$$
therefore
$$EG - F^2 = (1+f_x^2)(1+f_y^2) - (f_xf_y)^2 = 1+f_x^2 +f_y^2,$$
agreeing with the calculus definition, except that this is more general than for graphs.
