1. Given differential equation

$y\,dx + \left(x + x^2y\right)dy = 0$

We rewrite it in a form involving $\dfrac{dx}{dy}$:

$y\frac{dx}{dy} + x + x^2y = 0$

so,

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

That is,

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

2. Recognize Bernoulli-type equation in $x$ as a function of $y$

This is of the form

$\frac{dx}{dy} + P(y)x = Q(y)x^2$

Let

$v = x^{-1} = \frac{1}{x}$

Then

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

Using

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

multiply throughout by $-\dfrac{1}{x^2}$:

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

So,

$\frac{dv}{dy} - \frac{1}{y}v = 1$

3. Solve the linear differential equation

$\frac{dv}{dy} - \frac{1}{y}v = 1$

Here,

$P(y) = -\frac{1}{y}$

Integrating factor:

$IF = e^{\int -\frac{1}{y}dy} = e^{-\ln y} = \frac{1}{y}$

Multiplying the equation by $\dfrac{1}{y}$:

$\frac{1}{y}\frac{dv}{dy} - \frac{1}{y^2}v = \frac{1}{y}$

Left side is:

$\frac{d}{dy}\left(\frac{v}{y}\right) = \frac{1}{y}$

Integrate:

$\frac{v}{y} = \int \frac{1}{y}dy = \ln y + C$

Thus,

$v = y(\ln y + C)$

Since $v = \dfrac{1}{x}$,

$\frac{1}{x} = y(\ln y + C)$

Divide by $y$:

$\frac{1}{xy} = \ln y + C$

Rearrange:

$\frac{1}{xy} - \ln y = C$

or equivalently,

$-\frac{1}{xy} + \ln y = C$

4. Match with options

The obtained solution is

$-\frac{1}{xy} + \ln y = C$

This matches Option B.

5. Verification of options

• A: $\log y = Cx$ — not the derived general solution.
• B: $-\dfrac{1}{xy} + \log y = C$ — correct.
• C: $\dfrac{1}{xy} + \log y = C$ — wrong sign.
• D: $-\dfrac{1}{xy} = C$ — incomplete; misses the $\log y$ term.

Hence, the correct answer is B.