The curve satisfying the differential equation, $$ydx - (x + 3y^2)dy = 0$$ and passing through the point $$(1, 1)$$ also passes through the point:

We are given the differential equation

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

and we know that the required curve passes through the point $$(1,1).$$ Our task is to integrate the differential equation, use the given point to find the constant of integration, and then check which of the four points also satisfies the obtained relation.

First, we isolate $$dx$$ in terms of $$dy$$ because that arrangement converts the equation into a standard linear form in the variable $$x$$ with respect to $$y$$:

$$y\,dx = \bigl(x + 3y^2\bigr)\,dy$$

Dividing both sides by $$y\,dy$$ (and assuming $$y \neq 0$$ on the curve, which is true for all the listed points), we get

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

Rewriting,

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

This is a first-order linear ordinary differential equation of the form

$$\frac{dx}{dy} + P(y)\,x = Q(y),$$

where here $$P(y) = -\dfrac{1}{y}$$ and $$Q(y) = 3y.$$

The standard formula says that the integrating factor (I.F.) for such an equation is

$$\text{I.F.} = e^{\int P(y)\,dy}.$$

Calculating that integral, we have

$$\int P(y)\,dy = \int\!\left(-\frac{1}{y}\right)\,dy = -\ln y.$$

Therefore

$$\text{I.F.} = e^{-\ln y} = y^{-1}.$$

We now multiply every term of the differential equation by this integrating factor $$y^{-1}$$:

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

The left side is purposely arranged so that it becomes the derivative of a product. Indeed, notice that

$$\frac{d}{dy}\!\Bigl(x\,y^{-1}\Bigr) = y^{-1}\,\frac{dx}{dy} + x\,\frac{d}{dy}\,(y^{-1}) = y^{-1}\,\frac{dx}{dy} - x\,y^{-2},$$

which is exactly the combination we obtained. Hence we can write

$$\frac{d}{dy}\!\Bigl(x\,y^{-1}\Bigr) = 3.$$

We integrate both sides with respect to $$y$$:

$$x\,y^{-1} = \int 3\,dy = 3y + C,$$

where $$C$$ is the constant of integration.

Multiplying by $$y$$ gives us the explicit relation between $$x$$ and $$y$$ on the required curve:

$$x = y\,(3y + C) = 3y^2 + C\,y.$$

Now we impose the condition that the curve passes through the point $$(1,1).$$ Substituting $$x = 1$$ and $$y = 1$$ into the relation, we have

$$1 = 3(1)^2 + C(1) \quad\Longrightarrow\quad 1 = 3 + C \quad\Longrightarrow\quad C = -2.$$

Hence the particular integral curve is described by the equation

$$x = 3y^2 - 2y.$$

We must now check which of the four proposed points satisfies this equation.

Option A: $$\left(\dfrac14,\,-\dfrac12\right)$$
For $$y = -\dfrac12$$ we compute $$x = 3\left(-\dfrac12\right)^{\!2} - 2\left(-\dfrac12\right) = 3\left(\dfrac14\right) + 1 = \dfrac34 + 1 = \dfrac74.$$ That is $$x = \dfrac74 \neq \dfrac14,$$ so the point does not lie on the curve.

Option B: $$\left(-\dfrac13,\,\dfrac13\right)$$
For $$y = \dfrac13$$ we compute $$x = 3\left(\dfrac13\right)^{\!2} - 2\left(\dfrac13\right) = 3\left(\dfrac19\right) - \dfrac23 = \dfrac13 - \dfrac23 = -\dfrac13.$$ Here $$x = -\dfrac13,$$ exactly matching the given point. So this point **does** lie on the curve.

Option C: $$\left(\dfrac14,\,\dfrac12\right)$$
For $$y = \dfrac12$$ we compute $$x = 3\left(\dfrac12\right)^{\!2} - 2\left(\dfrac12\right) = 3\left(\dfrac14\right) - 1 = \dfrac34 - 1 = -\dfrac14,$$ which is not equal to $$\dfrac14,$$ so this point is not on the curve.

Option D: $$\left(\dfrac13,\,-\dfrac13\right)$$
For $$y = -\dfrac13$$ we compute $$x = 3\left(-\dfrac13\right)^{\!2} - 2\left(-\dfrac13\right) = 3\left(\dfrac19\right) + \dfrac23 = \dfrac13 + \dfrac23 = 1,$$ and clearly $$1 \neq \dfrac13,$$ so this point is also not on the curve.

Only Option B satisfies the integrated relation obtained from the differential equation. Therefore the curve passing through $$(1,1)$$ also passes through $$\left(-\dfrac13,\dfrac13\right).$$

Hence, the correct answer is Option 2.