= Solution
Separate the variables in this <first-order ordinary differential equation>:
$$
\frac{y}{y^2+2}\,dy=\frac{x}{x^2-3}\,dx.
$$
Integration gives $\tfrac12\ln(y^2+2)=\tfrac12\ln|x^2-3|+C$, so $y^2+2=K|x^2-3|$. On the interval containing the prescribed point $x=3$, we have $x^2-3>0$. The <initial condition> $y(3)=4$ therefore gives $18=6K$, hence $K=3$. Choosing the positive <square root> to match that <initial condition> yields
$$
\boxed{y(x)=\sqrt{3x^2-11}.}
$$
The maximal interval containing $3$ on which this solves the explicit <ordinary differential equation> is $\boxed{x>\sqrt{11/3}}$. At the left endpoint $y=0$, the original denominator vanishes and the <derivative> becomes unbounded. Directly, $y'=3x/y$ and $x(2+y^2)/[y(x^2-3)]=3x/y$, verifying the equation throughout this interval.
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