 Solution

Near $y=1$, the right side is continuously differentiable in $y$, with derivative $1/(4y^{3/4})$ bounded on, for example, $1/2\le y\le 3/2$. It therefore satisfies a local Lipschitz condition; it is also bounded on an appropriate bounded neighborhood. The existence and uniqueness result applies there. Separation of variables givesThe unique extension to the whole real axis is $u(x)=[\max \{1+3x/4,0\}{]}^{4/3}$. It joins zero with zero derivative at $x=-4/3$. To see uniqueness of this extension, $u^{\prime }\ge 0$ makes every solution nondecreasing. Before it reaches zero it cannot be positive, while a negative interval would have $u^{\prime }=0$ and could not continuously reach zero. Thus it must be zero for $x\le -4/3$; the positive part is fixed by the separated formula.