I'm working through a question where the differential equation is
$$ y^2(y'^2 -1)(3y'^2 +1) = c, \;\;\; y(0) = 0 $$
and the answer proceeds with two cases
(1) $c=0 \implies y(x)=0 \vee y(x) = \pm x$
(2) $c\neq0 \implies \lim_{x\to 0} y^2y'^4 = c/3$
I'm fine with (1), but I can't see how to get (2). I've tried multiplying out the brackets and eliminating terms involving $y$ only, but that doesn't get the answer so far as I can see. I'm not very familiar with how to manipulate limits, especially in the context of differential equations, and would appreciate some help.
Near $x=0$, $y$ is very small, so $y'$ must be large if $c\neq0$. If $y'$ is large then the terms in brackets will approximate to $3y'^4$.