1. Given differential equation

$y^2\,dx+\left(x-\frac1y\right)dy=0$

We are given the condition:

$y=1 \text{ when } x=1$

and we need to find $x$ when $y=2$.

2. Treat $x$ as a function of $y$

Since the equation contains $dx$ and $dy$, write it as:

$y^2\frac{dx}{dy}+x-\frac1y=0$

So,

$y^2\frac{dx}{dy}+x=\frac1y$

Divide by $y^2$:

$\frac{dx}{dy}+\frac{1}{y^2}x=\frac{1}{y^3}$

This is a linear differential equation in $x$ with independent variable $y$.

3. Find the integrating factor

The standard form is:

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

with

$P(y)=\frac1{y^2}, \qquad Q(y)=\frac1{y^3}$

Hence the integrating factor is:

$\text{I.F.}=e^{\int \frac1{y^2}dy}=e^{-1/y}$

because

$\int y^{-2}dy=-y^{-1}=-\frac1y$

4. Multiply throughout by the integrating factor

$e^{-1/y}\frac{dx}{dy}+\frac{1}{y^2}e^{-1/y}x=\frac{1}{y^3}e^{-1/y}$

The left side becomes:

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

So,

$xe^{-1/y}=\int \frac{1}{y^3}e^{-1/y}\,dy + C$

5. Evaluate the integral

We compute:

$I=\int \frac{1}{y^3}e^{-1/y}\,dy$

Let

$t=-\frac1y$

Then

$\frac{dt}{dy}=\frac1{y^2} \quad \Rightarrow \quad dt=\frac1{y^2}dy$

Now,

$\frac1{y^3}dy=\frac1y\cdot \frac1{y^2}dy$

Since $t=-1/y$, we have $1/y=-t$. Thus

$I=\int (-t)e^t\,dt=-\int te^t\,dt$

Using integration by parts,

$\int te^t\,dt=e^t(t-1)$

Therefore,

$I=-e^t(t-1)=e^t(1-t)$

Substitute back $t=-1/y$:

$I=e^{-1/y}\left(1+\frac1y\right)$

Hence,

$xe^{-1/y}=e^{-1/y}\left(1+\frac1y\right)+C$

Multiply by $e^{1/y}$:

$x=1+\frac1y+Ce^{1/y}$

6. Use the initial condition

Given $x=1$ when $y=1$:

$1=1+1+Ce$

$1=2+Ce$

$Ce=-1 \Rightarrow C=-\frac1e$

So the solution is:

$x=1+\frac1y-\frac1e e^{1/y}$

$x=1+\frac1y-e^{\frac1y-1}$

7. Find $x$ when $y=2$

Substitute $y=2$:

$x=1+\frac12-e^{\frac12-1}$

$x=\frac32-e^{-1/2}$

$x=\frac32-\frac1{\sqrt e}$

8. Compare with the options

This matches:

$\boxed{\frac32-\frac1{\sqrt e}}$

So the correct option is A.

9. Verification with stored answer

Stored correct answer: A

Our derived answer: A

Hence, they agree.