This is a somewhat silly problem, since the solution can be found without the calculus of variations: The functional is non-negative and is zero when $y'=0$, so the solution subject to $y(0)=0$ is $y(x)\equiv0$.
To do this with the calculus of variations, you can either use the natural boundary condition
which yields $y'(1)=0$, or you can solve the problem with $y(1)$ unspecified and then extremize the functional with respect to the parameter of the solution. In either case, the Euler-Lagrange equation is $2y''+y'^2=0$ (you got the sign wrong), and as Peter pointed out this can be solved by substituting $u=y'$. The result is $y(x)=2\log(x+c)+c'$. The boundary condition $y(0)=0$ yields $c'=-2\log c$ and thus $y(x)=2\log (x/c+1)$.
Now applying the natural boundary condition gives $c=\infty$ and thus $y(x)\equiv0$; or you can calculate the functional, $J(c)=4/c^2$, and again $c=\infty$ gives the extremum.