1. Rewrite the differential equation

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

Since $x=x(y)$, divide by $dy$: $y^2\frac{dx}{dy}+x-\frac{1}{y}=0.$

So, $y^2\frac{dx}{dy}+x=\frac{1}{y}.$

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

This is a linear differential equation in $x$ as a function of $y$.

2. Find the integrating factor

For $\frac{dx}{dy}+P(y)x=Q(y),$ we have $P(y)=\frac{1}{y^2}.$

Hence the integrating factor is $\text{I.F.}=e^{\int \frac{1}{y^2}dy}=e^{-1/y}.$

3. Multiply throughout by the integrating factor

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

The left-hand side is $\frac{d}{dy}\left(xe^{-1/y}\right)=\frac{e^{-1/y}}{y^3}.$

So we need to integrate: $xe^{-1/y}=\int \frac{e^{-1/y}}{y^3}\,dy + C.$

4. Evaluate the integral

Let $t=-\frac{1}{y}.$ Then $dt=\frac{1}{y^2}dy.$ A better direct substitution is:

Let $u=-\frac{1}{y} \implies du=\frac{1}{y^2}dy.$

Now, $\frac{1}{y^3}dy=\frac{1}{y}\cdot \frac{1}{y^2}dy=-\!u\,du$ since $u=-1/y \Rightarrow 1/y=-u$.

Thus, $\int \frac{e^{-1/y}}{y^3}dy=\int e^u(-u)\,du=-\int ue^u\,du.$

Using integration by parts, $\int ue^u\,du=e^u(u-1).$ Therefore, $-\int ue^u\,du=-e^u(u-1)=e^u(1-u).$

Substitute back $u=-1/y$: $\int \frac{e^{-1/y}}{y^3}dy=e^{-1/y}\left(1+\frac{1}{y}\right).$

Hence, $xe^{-1/y}=e^{-1/y}\left(1+\frac{1}{y}\right)+C.$

Multiplying by $e^{1/y}$, $x=1+\frac{1}{y}+Ce^{1/y}.$

5. Use the initial condition

Given $x(1)=1$: $1=1+1+Ce.$ So, $Ce=-1 \implies C=-\frac{1}{e}.$

Therefore, $x(y)=1+\frac{1}{y}-e^{1/y-1}.$

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

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

7. Compare with options

The value is $3-e,$ which matches Option D.