NIMCET 2019 Question Paper with solutions

Given: $\mathit{y}\mathit{d}\mathit{x}-\left(\mathit{x}+3{\mathit{y}}^2\right)\mathit{d}\mathit{y}=0$
$\Rightarrow \mathit{y}\mathit{d}\mathit{x}-\mathit{x}\mathit{d}\mathit{y}-3{\mathit{y}}^2\mathit{d}\mathit{y}=0$
$\Rightarrow \mathit{y}\mathit{d}\mathit{x}-\mathit{x}\mathit{d}\mathit{y}=3{\mathit{y}}^2\mathit{d}\mathit{y}$
$\Rightarrow \frac{\mathit{y}\mathit{d}\mathit{x}-\mathit{x}\mathit{d}\mathit{y}}{{\mathit{y}}^2}=3\mathit{d}\mathit{y}$
$\Rightarrow \mathit{d}\left(\frac{\mathit{x}}{\mathit{y}}\right)=3\mathit{d}\mathit{y}$
Integrating both sides, we get
$\Rightarrow \int \mathit{d}\left(\frac{\mathit{x}}{\mathit{y}}\right)=\int 3\mathit{d}\mathit{y}$
$\Rightarrow \frac{\mathit{x}}{\mathit{y}}=3\mathit{y}+\mathit{c}$
$\Rightarrow \mathit{x}=3{\mathit{y}}^2+\mathit{c}\mathit{y}.........\left(\mathit{i}\right)$
Curve passing through the point $\left(1,1\right)$
$\Rightarrow 1=3+\mathit{c}$
$\therefore \mathit{c}=-2$
Put the value of $\mathit{c}$ in equation 1 st we get,
$\Rightarrow \mathit{x}=3{\mathit{y}}^2-2\mathit{y}$
Checking from options:
This equation is satisfied by point $\left(-1∕3,1∕3\right)$
Hence option 3 is correct.