Find the extremal $y(x) $ for $ \int^{3}_{0}(x(y')^3 -3y(y')^2)dx $ when $y(3) = 1$ and $y(0) = 7$ 
Find the extremal $y(x)$ for:
$$ \int^{3}_{0}(x(y')^3 -3y(y')^2)dx  $$ 
  when $y(3) = 1$ and $y(0) = 7$.

I know I need to use the following Euler-Lagrange connection:
$$\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial u_x{}}\right)  + 
\frac{\partial}{\partial y}\left(\frac{\partial F}{\partial u_y{}}\right) -\frac{\partial F}{\partial u}=0.$$
My question is, when plugging $F(x,x',y,y')$ into Euler-Lagrange, how do I calculate 
$$\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial u_x{}}\right)~?$$
Is it zero since there are no terms that contain $u_x$? And is $\frac{\partial F}{\partial u}=0$ as well since u doesn't appear explicitly?
 A: You are using the wrong Euler Lagrange equation. Here $x$ is an independent variable and $y$ is a function of $x$ so you have $F = F(x,y,y')$. The version given is for when you have a function $u$ of two independent variables $x,y$ so $F = F(x,y,u,u_x,u_y)$. Your E-L equation should look like $$\frac{\partial F}{\partial y} = \frac{d}{dx}\left( \frac{\partial F}{\partial y'} \right)$$ where $$F(x,y,y') = x(y')^3 - 3y(y')^2.$$ Thus the equation is $$-3(y')^2 = \frac{d}{dx}( 3x(y')^2 - 6yy') = 3(y')^2 + 6xy' - 6(y')^2 - 6yy''$$ which simplifies to $$yy'' = xy'.$$ 
A: OP conflates different notations for the dependent and independent variables in the variational problem, as already explained in User8128's answer. In this answer, we will concentrate on solving the variational problem itself. The functional
$$ F[y]~=~\int_0^3  \! \mathrm{d}x~(xy^{\prime}-3y)y^{\prime 2}\tag{A}$$
has Euler-Lagrange equation 
$$ (xy^{\prime}-y)y^{\prime\prime}~=~0.\tag{B}$$ 
Note the factorized form of eq. (B). The full solution to eq. (B) is an affine function
$$ y(x)~=~ax+b.\tag{C}$$
If we implement the boundary conditions
$$ y(0)~=~7\qquad\text{and}\qquad y(3)~=~1, \tag{D} $$
the solution becomes 
$$ y(x)~=~7-2x. \tag{E} $$
