= Solution
{c}
<Separation of variables> gives
$$
\frac{y\,dy}{3y^2+1}=\frac{x\,dx}{x^2-1},\qquad
\frac16\log(3y^2+1)=\frac12\log(x^2-1)+C.
$$
Here $x>1$ makes the argument of the second <logarithm> positive. Substituting the <initial condition> yields $3y^2+1=4(x^2-1)^3$. The positive branch is therefore
$$
\boxed{y(x)=\sqrt{\frac{4(x^2-1)^3-1}{3}},\qquad x>\sqrt{1+4^{-1/3}}.}
$$
This is the maximal connected <interval> containing $\sqrt2$ on which this branch solves the original <differential equation>. At its left endpoint $y=0$, the original right-hand side has a zero denominator, so one cannot extend the solution through that point merely by continuing the squared equation.
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