If a curve $$y = f(x)$$ passes through the point $$(1, -1)$$ and satisfies the differential equation, $$y(1 + xy)dx = x\,dy$$, then $$f\left(-\frac{1}{2}\right)$$ is equal to

We are given the differential equation $$y(1+xy)\,dx = x\,dy$$ and the information that the required curve passes through the point $$(1,-1)$$. Our goal is to find the value of the function at $$x=-\dfrac12$$, that is, to compute $$f\!\left(-\dfrac12\right)$$.

First we bring the equation into the usual $$\dfrac{dy}{dx}$$ form. Dividing both sides by $$dx$$ we have

$$y(1+xy)=x\,\dfrac{dy}{dx}.$$

Now we isolate $$\dfrac{dy}{dx}$$ by dividing by $$x$$:

$$\dfrac{dy}{dx}= \dfrac{y(1+xy)}{x}= \dfrac{y}{x}+y^{2}.$$

So we obtain the first-order ordinary differential equation

$$\dfrac{dy}{dx}-y^{2}= \dfrac{y}{x}. \quad -(1)$$

A very convenient substitution for equations of the form $$\dfrac{dy}{dx}-y^{2}=\dots$$ is to set $$v=\dfrac1y$$, because the derivative of $$\dfrac1y$$ naturally cancels the $$y^{2}$$ term. Let us therefore put

$$v=\dfrac1y.$$

Differentiating both sides with respect to $$x$$ gives

$$\dfrac{dv}{dx}= -\dfrac1{y^{2}}\dfrac{dy}{dx}.$$

We now substitute $$\dfrac{dy}{dx}= -y^{2}\dfrac{dv}{dx}$$ into equation (1):

$$-y^{2}\dfrac{dv}{dx}-y^{2}= \dfrac{y}{x}.$$

Multiplying every term by $$-1$$ produces

$$y^{2}\dfrac{dv}{dx}+y^{2}= -\dfrac{y}{x}.$$

Next we divide throughout by $$y^{2}$$ (remembering that $$y\neq 0$$ along the solution curve):

$$\dfrac{dv}{dx}+1= -\dfrac1{xy}.$$

But $$\dfrac1y=v,$$ so $$-\dfrac1{xy}=-\dfrac{v}{x}$$, hence

$$\dfrac{dv}{dx}+1=-\dfrac{v}{x}.$$

We collect like terms and arrange in the standard linear form

$$\dfrac{dv}{dx}+\dfrac{v}{x}= -1. \quad -(2)$$

Equation (2) is a first-order linear differential equation in $$v(x)$$. For such equations we use the integrating-factor method. A linear equation has the standard form $$\dfrac{dv}{dx}+P(x)\,v = Q(x)$$. Here $$P(x)=\dfrac1x$$ and $$Q(x)=-1.$$ The integrating factor $$\mu(x)$$ is defined by the formula

$$\mu(x)=e^{\displaystyle\int P(x)\,dx}.$$

We therefore compute the integrating factor:

$$\mu(x)=e^{\displaystyle\int \frac1x\,dx}=e^{\ln x}=x.$$

Multiplying every term of equation (2) by this integrating factor $$x$$ we get

$$x\,\dfrac{dv}{dx}+v = -x.$$

The left-hand side is now the perfect derivative of the product $$xv$$, because

$$\dfrac{d}{dx}(xv)=x\,\dfrac{dv}{dx}+v.$$

Therefore we can rewrite the equation as

$$\dfrac{d}{dx}(xv)= -x.$$

To find $$v(x)$$ we integrate both sides with respect to $$x$$:

$$\int \dfrac{d}{dx}(xv)\,dx = \int -x\,dx,$$

so

$$xv= -\dfrac{x^{2}}2 + C,$$

where $$C$$ is the constant of integration. Solving for $$v$$ gives

$$v= -\dfrac{x}{2}+\dfrac{C}{x}.$$

Recalling that $$v=\dfrac1y$$, we obtain

$$\dfrac1y= -\dfrac{x}{2}+\dfrac{C}{x}.$$

We bring the right-hand side over a common denominator $$x$$:

$$\dfrac1y=\dfrac{-x^{2}/2+C}{x}.$$

Taking the reciprocal to solve for $$y$$, we get

$$y=\dfrac{x}{-x^{2}/2+C}.$$

To make the formula look cleaner, multiply numerator and denominator by $$2$$; let us also rename the constant $$2C$$ by a single letter $$K$$, i.e. set $$K=2C$$. Then

$$y=\dfrac{2x}{K-x^{2}}.$$

We now use the given point $$(1,-1)$$ to determine the value of the constant $$K$$. Substituting $$x=1$$ and $$y=-1$$ into the above expression yields

$$-1=\dfrac{2\cdot 1}{K-1},$$

so

$$K-1 = -2,$$

and hence

$$K=-1.$$

Substituting $$K=-1$$ back into the expression for $$y$$ we obtain the explicit form of the required function:

$$y=f(x)=\dfrac{2x}{-1-x^{2}} = -\dfrac{2x}{1+x^{2}}.$$

With the function determined, we can now evaluate it at $$x=-\dfrac12$$. First compute the numerator:

$$-2x = -2\!\left(-\dfrac12\right)=1.$$

Next compute the denominator:

$$1+x^{2}=1+\left(-\dfrac12\right)^{2}=1+\dfrac14=\dfrac54.$$

Putting numerator and denominator together, we find

$$f\!\left(-\dfrac12\right)=\dfrac{1}{\dfrac54}=1\cdot \dfrac45=\dfrac45.$$

Thus the required value is $$\dfrac45$$, which corresponds to the second option listed.

Hence, the correct answer is Option B.