= Solution
Near $y=1$, the right side is continuously differentiable in $y$, with <derivative> $1/(4y^{3/4})$ bounded on, for example, $1/2\le y\le3/2$. 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> gives
$$
u^{3/4}=1+\frac{3x}{4},\qquad
\boxed{u(x)=\left(1+\frac{3x}{4}\right)^{4/3}\quad(x>-4/3).}
$$
The 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, $u'\ge0$ makes every solution nondecreasing. Before it reaches zero it cannot be positive, while a negative interval would have $u'=0$ 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.
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