Euler Lagrange equation of $J[y]=\int_0^1 (yy')^2dx$ subject to the constraint $\int_0^1 y^2 dx =3$ 
Among all the admissible functions $y = y(x)$, find those that extremise the functional
  $$J[y] = \int_0^1 (yy')^2dx$$ 
  subject to the constraint $\int_0^1 y^2 dx =3$ and the boundary conditions $y(0)=1, y(1)=2$.

Am I correct in saying denoting $F=(yy')^2 + \lambda y^2$. Thus the extrema corresponding to this problem are the extrema for the functional,
$$J[y] =\int_0^1 F dx$$
and therefore they are solutions of the Euler-Lagrange equation
$$\lambda y -y^2y''-(yy')^2=0.$$
Could some clarify whether my Euler-Lagrange equation is correct. I am pretty confused as to where to go from here. Any help would be grand.
 A: 
Am I correct in saying denoting $F=(yy')^2 + \lambda y^2$. Thus the extrema corresponding to this problem are the extrema for the functional,
  $$J[y] = \int_0^1 F dx.$$

You are correct. If we wanna minimize a functional $J[y]$ subject to a constraint $H[y]=\text{const}$, then the new functional we want to minimize is 
$$
J_{\lambda}[y]:= J[y]+\lambda H[y] = \int^1_0 \left((yy')^2+\lambda y^2\right) dx.\tag{1}
$$


Therefore they are solutions of the Euler-Lagrange equation
  $$\lambda y -y^2y''-(yy')^2=0$$
  Could some clarify whether my Euler-Lagrange equation is correct. 

Upon my checking, I think this is not correct, equation $(\star)$ in blue color is what I obtained. 
Instead of plugging the formula, I still prefer the old school way of deriving the E-L equation, aka the first variation of the functional in (1) is zero when the extremum is attained at $y$:
$$
\lim_{\epsilon\to 0} \frac{d}{d\epsilon} J_{\lambda}[y+\epsilon v] = 0,\\
\text{for any smooth test function } v \text{ that does not change the boundary value of } y.
$$
This is to say $v=0$ on both end points.
Firstly computing $J_{\lambda}[y+\epsilon v]$ yields:
$$
\int^1_0 \left((y+\epsilon v)^2(y'+\epsilon v')^2+ \lambda (y+\epsilon v)^2\right) dx.
$$
Taking derivative w.r.t. $\epsilon$ yields:
$$
\int^1_0 \left(2(y+\epsilon v)v(y'+\epsilon v')^2+2(y+\epsilon v)^2 (y'+\epsilon v') v' + 2\lambda (y+\epsilon v) v\right) dx.
$$
Letting $\epsilon\to 0$ and setting the integral to be zero yield:
$$
\int^1_0 \left(  y(y')^2v +  \color{red}{y^2  y' v'} +  \lambda  y  v\right) dx = 0.
$$
Integrating by parts on the red term and using $v=0$ on the end points:
$$
\int^1_0 \left(  y(y')^2v - 2y(y')^2 v - y^2y''v +  \lambda  y  v\right) dx = 0.
$$
By fundamental theorem of calculus of variation:
$$
\color{blue}{y(y')^2 + y^2y'' =\lambda y}.\tag{$\star$} 
$$
Suppose $y\neq 0$ for any $x$, we have
$$
(y')^2 + y y'' =\lambda y
\implies (y^2)'' = 2\lambda.
$$
Solving this yields:
$$
y =\pm \sqrt{\lambda x^2 + ax + b}.
$$
The constraint $\int^1_0 y^2\,dx = 3$, the boundary condition $y(0)=1$, $y(1)=2$ will pin down the parameters, and we reached the final answer:
$$
\boxed{y = \sqrt{-3x^2+6x+1}}.
$$
