The particular solution of the differential equation \(\left(1+y^2\right) \mathrm{d} x-x y \mathrm{~d} y=0\) at \(x=1, y=0\), represents

\(\begin{array}{ll}
& \left(1+y^2\right) \mathrm{d} x-x y \mathrm{~d} y=0 \\
\therefore \quad & \left(1+y^2\right) \mathrm{d} x=x y \mathrm{~d} y \\
\therefore \quad & \frac{1}{x} \mathrm{~d} x=\frac{y \mathrm{~d} y}{1+y^2}
\end{array}\)
Integrating both sides, we get
\(\begin{aligned}
& \int \frac{1}{x} \mathrm{~d} x=\int \frac{y}{1+y^2} \mathrm{~d} y \\
& \log x=\frac{1}{2} \log \left(1+y^2\right)+\mathrm{c}
\end{aligned}\)
\(\text { At } x=1, y=0...[Given]\)
\(\begin{array}{ll}
\therefore & 0=\frac{1}{2} \log (1+0)+\mathrm{c} \\
\therefore & \quad \mathrm{c}=0 \\
\therefore & \log x=\frac{1}{2} \log \left(1+y^2\right) \\
\therefore & x^2=1+y^2 \\
\therefore & x^2-y^2=1,
\end{array}\)
Which is a rectangular hyperbola.