Answer
$y=\sqrt{4x^2+3}$
Work Step by Step
$y(1+x^2)y'-x(1+y^2)=0$
$y(1+x^2)y'=x(1+y^2)$
$\frac{yy'}{1+y^2}=\frac{x}{1+x^2}$
$\int \frac{y}{1+y^2}dy=\int \frac{x}{1+x^2}dx$
$\frac{1}{2}\ln{(1+y^2)}=\frac{1}{2}\ln{(1+x^2)}+C$
$\ln{(1+y^2)}=\ln{(1+x^2)}+C$
$e^{\ln{(1+y^2)}}=1+y^2=e^{\ln{(1+x^2)}+C}=e^Ce^{\ln{(1+x^2)}}$
Let $k=e^C$ and solve for $y$
$y=\sqrt{k(1+x^2)-1}$.
Now applying our initial condition we get
$\sqrt{3}=\sqrt{k(1+0^2)-1}=\sqrt{k-1}$
Therefore k=4, and our particular solution is
$y=\sqrt{4x^2+3}$
