The curve satisfying the differential equation \(y \mathrm{~d} x-\left(x+3 y^2\right) \mathrm{dy}=0\) and passing through the point \((1,1)\) also passes through the point

\(\begin{aligned} & y \mathrm{~d} x-\left(x+3 y^2\right) \mathrm{d} y=0 \\ & \Rightarrow y \mathrm{~d} x=\left(x+3 y^2\right) \mathrm{d} y \\ & \Rightarrow \frac{\mathrm{~d} x}{\mathrm{~d} y}=\frac{x+3 y^2}{y} \\ & \Rightarrow \frac{\mathrm{~d} x}{\mathrm{~d} y}=\frac{x}{y}+3 y \\ & \Rightarrow \frac{\mathrm{~d} x}{\mathrm{~d} y}-\left(\frac{1}{y}\right) x=3 y\end{aligned}\)
Which is a linear equation
\(\therefore \quad \mathrm{IF}=\mathrm{e}^{\int \frac{-1}{y} \mathrm{~d} y}=\mathrm{e}^{-\log y}=\frac{1}{y}\)
\(\therefore \quad\) The required solution is
\(x \frac{1}{y}=\int 3 y \times \frac{1}{y} \mathrm{~d} y+\mathrm{c}\)
\(\begin{aligned}
\therefore \quad & \frac{x}{y}=3 y+c \\
& \Rightarrow x=3 y^2+c y
\end{aligned}\)
Curve passes through \((1,1)\)
\(\begin{aligned}
& \Rightarrow 1=3+c \\
& \Rightarrow c=-2
\end{aligned}\)
Equation (i) becomes,
\(x=3 y^2-2 y\)
Option (D) i.e., \(\left(\frac{-1}{3}, \frac{1}{3}\right)\)
Satisfies above equation.