We have,
$$
\begin{aligned}
& \frac{d y}{d x}+\frac{x}{y}\left(\frac{x^2+y^2-1}{2\left(x^2+y^2\right)+1}\right)=0 \\
\Rightarrow & 2 y\left(x^2+y^2\right) d y+y d y+x\left(x^2+y^2\right) d x-x d x=0 \\
\Rightarrow & 2 y\left(x^2+y^2+2-2\right) d y+y d y+ \\
& x\left(x^2+y^2+2-2\right) d x-x d x=0 \\
\Rightarrow & 2 y\left(x^2+y^2+2\right) d y-4 y d y+y d y \\
\Rightarrow & 2 y\left(x^2+y^2+2\right) d y+x\left(x^2+y^2+2\right) d x-3 x d x=0 \\
\Rightarrow & 2 y d y+x d x=3\left(\frac{x d x+y d y}{x^2+y^2+2}\right) \quad=3 x d x+3 y d y \\
\Rightarrow & 2 y d y+x d x=\frac{3}{2}\left(\frac{2 x d x+2 y d y}{x^2+y^2+2}\right) \\
\Rightarrow & 2 y d y+x d x=\frac{3}{2} d\left(\log \left(x^2+y^2+2\right)\right.
\end{aligned}
$$
On integrating both sides, we get
$$
\begin{aligned}
& 2 y^2+x^2=3 \log \left(x^2+y^2+2\right)+c \\
& x^2+2 y^2-3 \log \left|x^2+y^2+2\right|=c
\end{aligned}
$$