Given differential equation

$y\,dx-(x+3y^2)\,dy=0$

Rewrite it by treating $x$ as a function of $y$:

$y\frac{dx}{dy}=x+3y^2$

$\frac{dx}{dy}-\frac{1}{y}x=3y$

This is a linear differential equation in $x$ with independent variable $y$.

Find the integrating factor

The standard form is $\frac{dx}{dy}+P(y)x=Q(y)$ with $P(y)=-\frac{1}{y}, \qquad Q(y)=3y.$

So the integrating factor is

$\text{I.F.}=e^{\int -\frac{1}{y}\,dy}=e^{-\ln y}=\frac{1}{y}$

(up to a constant factor).

Multiply the equation by the integrating factor

$\frac{1}{y}\frac{dx}{dy}-\frac{x}{y^2}=3$

The left side is

$\frac{d}{dy}\left(\frac{x}{y}\right)=3$

Hence,

$\frac{x}{y}=3y+C$

$x=3y^2+Cy$

Use the point $(1,1)$

Since the curve passes through $(1,1)$,

$1=3(1)^2+C(1)$

$1=3+C$

$C=-2$

Therefore the curve is

$x=3y^2-2y$

Check the options

We test each point in $x=3y^2-2y$

• Option A: $\left(\frac14,-\frac12\right)$ $3\left(\frac14\right)-2\left(-\frac12\right)=\frac34+1=\frac74\neq \frac14$ So A is false.

• Option B: $\left(-\frac13,\frac13\right)$ $3\left(\frac{1}{9}\right)-2\left(\frac13\right)=\frac13-\frac23=-\frac13$ This matches $x=-\frac13$. So B is true.

• Option C: $\left(\frac13,-\frac13\right)$ $3\left(\frac{1}{9}\right)-2\left(-\frac13\right)=\frac13+\frac23=1\neq \frac13$ So C is false.

• Option D: $\left(\frac14,\frac12\right)$ $3\left(\frac14\right)-2\left(\frac12\right)=\frac34-1=-\frac14\neq \frac14$ So D is false.

Conclusion

The required point is $\left(-\frac13,\frac13\right)$

Hence the correct option is B.