1. Rewrite the differential equation

Given $x^2\,dy + \left(y - \frac{1}{x}\right)dx = 0.$

Divide by $dx$: $x^2\frac{dy}{dx} + y - \frac{1}{x} = 0.$ So, $x^2\frac{dy}{dx} = \frac{1}{x} - y,$ $\frac{dy}{dx} + \frac{1}{x^2}y = \frac{1}{x^3}.$

This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x),$ with $P(x)=\frac{1}{x^2}, \qquad Q(x)=\frac{1}{x^3}.$

2. Find the integrating factor

The integrating factor is $\text{I.F.} = e^{\int \frac{1}{x^2}dx} = e^{-1/x}.$

3. Multiply the equation by the integrating factor

Multiplying throughout by $e^{-1/x}$, $e^{-1/x}\frac{dy}{dx} + \frac{1}{x^2}e^{-1/x}y = \frac{1}{x^3}e^{-1/x}.$

The left-hand side becomes $\frac{d}{dx}\left(ye^{-1/x}\right).$ Thus, $\frac{d}{dx}\left(ye^{-1/x}\right)=\frac{1}{x^3}e^{-1/x}.$

4. Integrate both sides

We need $ye^{-1/x} = \int \frac{1}{x^3}e^{-1/x}dx + C.$

Let $t=-\frac{1}{x} \implies dt=\frac{1}{x^2}dx.$ Then $\frac{1}{x^3}dx=\frac{1}{x}\cdot \frac{1}{x^2}dx = -t\,dt.$ So the integral becomes $\int \frac{1}{x^3}e^{-1/x}dx = \int (-t)e^t dt.$

Now, $\int t e^t dt = e^t(t-1),$ therefore $\int (-t)e^t dt = -e^t(t-1) = e^t(1-t).$

Substituting back $t=-\frac{1}{x}$, $\int \frac{1}{x^3}e^{-1/x}dx = e^{-1/x}\left(1+\frac{1}{x}\right).$

Hence, $ye^{-1/x} = e^{-1/x}\left(1+\frac{1}{x}\right) + C.$ Multiplying by $e^{1/x}$, $y = 1 + \frac{1}{x} + Ce^{1/x}.$

5. Use the initial condition $y(1)=1$

Substitute $x=1, y=1$: $1 = 1 + 1 + Ce,$ $Ce = -1,$ $C=-\frac{1}{e}.$

So the solution is $y = 1 + \frac{1}{x} - e^{1/x-1}.$

6. Find $y\left(\frac12\right)$

Substitute $x=\frac12$: $y\left(\frac12\right)=1+2-e^{2-1}=3-e.$

7. Match with the options

$y\left(\frac12\right)=3-e,$ which is Option D.