Variational integral problem with corners

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Consider the integral

$$ J(y) = \int_{x_1}^{x_2} \frac{y(x)}{(y'(x))^2} + (y'(x))^2dx\,, \ \ \ \ y'(x) \neq 0 $$

Show that for this problem corners are possible. (Hint: The corner conditions are satisfied if the substitution $y′_r(x_0) = −y′_l(x_0)$ is made.)

As the integrand $f=f(y(x),y'(x))$ the Euler equation reduces to

$$ f-y'\frac{df}{dy'}=constant=c1 $$

Which gives $$ \frac{3y}{(y')^2}-(y')^2=c1 $$

Any help from here would be much appreciated.