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Step 1: Write the differential equation in a convenient form
We start with the given equation:
$$y \bigl(1 + xy\bigr)\,dx = x\,dy.$$
Rearrange it to isolate the terms involving $dx$ and $dy$ in a helpful way:
$$y (1 + xy)\,dx - x\,dy = 0.$$
Step 2: Express in terms of a known derivative
Observe that
$$d\left(\frac{x}{y}\right) = \frac{y\,dx - x\,dy}{y^2}.$$
From the given differential equation, we can rearrange to match part of this pattern. Rewrite the left side in a form that features $(x\,dy - y\,dx)$:
Starting from
$$y (1 + xy)\,dx = x\,dy \quad\Longrightarrow\quad x\,dy - y\,dx = x^2 y^2 \, dx \;\;(\text{after appropriate rearrangements}).$$
Now divide both sides by $y^2$:
$$\frac{x\,dy - y\,dx}{y^2} = x\,dx.$$
Because
$$\frac{x\,dy - y\,dx}{y^2} = - \biggl(\frac{y\,dx - x\,dy}{y^2}\biggr)
= - \, d\!\Bigl(\frac{x}{y}\Bigr),$$
we get
$$- \, d\Bigl(\frac{x}{y}\Bigr) = x\,dx.$$
Step 3: Integrate both sides
Integrate term by term:
$$\int -\, d\!\Bigl(\frac{x}{y}\Bigr) = \int x\,dx.$$
The left side integrates to
$$-\, \frac{x}{y},$$
while the right side integrates to
$$\frac{x^2}{2} + C,$$
where $C$ is the constant of integration. Hence,
$$- \frac{x}{y} = \frac{x^2}{2} + C.$$
Step 4: Solve for $y$ using the initial condition
Rearrange to isolate $y$:
$$\frac{x}{y} = -\frac{x^2}{2} - C.$$
Or
$$-\frac{x}{y} = \frac{x^2}{2} + C.$$
We use the initial condition that the curve passes through $(1, -1)$, i.e. $y(1) = -1.$ Substitute $x = 1$ and $y = -1$:
$$- \frac{1}{-1} = \frac{1^2}{2} + C
\quad\Longrightarrow\quad 1 = \frac{1}{2} + C
\quad\Longrightarrow\quad C = \frac{1}{2}.$$
Thus,
$$- \frac{x}{y} = \frac{x^2}{2} + \frac{1}{2}
\quad\Longrightarrow\quad \frac{x}{y} = -\frac{x^2}{2} - \frac{1}{2}.$$
Solving for $y$ gives
$$y = \frac{-2x}{x^2 + 1}.$$
Step 5: Evaluate $f\bigl(-\tfrac12\bigr)$
We now substitute $x = -\tfrac12$ into the expression for $y$:
$$y = \frac{-\,2\left(-\frac12\right)}{\left(-\frac12\right)^2 + 1}
= \frac{1}{\frac{1}{4} + 1}
= \frac{1}{\frac{5}{4}}
= \frac{4}{5}.$$
Hence,
$$f\Bigl(-\tfrac12\Bigr) = \frac{4}{5}.$$
