Let $$y = y(x)$$ be the solution of the differential equation $$x^3 dy + (xy - 1) dx = 0$$, $$x > 0$$, $$y\left(\frac{1}{2}\right) = 3 - e$$. Then $$y(1)$$ is equal to

The differential equation is given in differential form as $$x^{3}\,dy + (x\,y - 1)\,dx = 0$$ with $$x \gt 0$$.

Divide by $$dx$$ to obtain the derivative explicitly:
$$x^{3}\,\frac{dy}{dx} + x\,y - 1 = 0$$

Isolate $$\dfrac{dy}{dx}$$:
$$\frac{dy}{dx} = -\frac{x\,y - 1}{x^{3}} = -\frac{y}{x^{2}} + \frac{1}{x^{3}}$$

Re-write in the standard linear form $$\dfrac{dy}{dx} + P(x)\,y = Q(x)$$:
$$\frac{dy}{dx} + \frac{1}{x^{2}}\,y = \frac{1}{x^{3}} \quad -(1)$$

Here $$P(x)=\dfrac{1}{x^{2}}$$ and $$Q(x)=\dfrac{1}{x^{3}}$$.

Integrating factor
For a linear ODE, the integrating factor is $$\mu(x)=\exp\!\left(\int P(x)\,dx\right)$$.
Evaluate the integral:
$$\int \frac{1}{x^{2}}\,dx = -\frac{1}{x}$$
Hence $$\mu(x)=e^{-1/x}$$.

Multiply equation $$(1)$$ by $$e^{-1/x}$$:
$$e^{-1/x}\,\frac{dy}{dx} + \frac{e^{-1/x}}{x^{2}}\,y = \frac{e^{-1/x}}{x^{3}}$$.

The left side is the derivative of $$y\,e^{-1/x}$$ (product rule in reverse):
$$\frac{d}{dx}\bigl(y\,e^{-1/x}\bigr) = \frac{e^{-1/x}}{x^{3}} \quad -(2)$$

Integrate both sides with respect to $$x$$:
$$y\,e^{-1/x} = \int \frac{e^{-1/x}}{x^{3}}\,dx + C.$$

Evaluating the integral
Let $$t = \frac{1}{x} \;\; \Longrightarrow \;\; dt = -\frac{1}{x^{2}}\,dx,$$ so $$dx = -x^{2}\,dt = -\frac{1}{t^{2}}\,dt$$ and $$\frac{1}{x^{3}} = t^{3}$$.
Thus

$$\int \frac{e^{-1/x}}{x^{3}}\,dx = \int e^{-t}\,t^{3}\left(-\frac{1}{t^{2}}\right)\,dt = -\int t\,e^{-t}\,dt.$$

Integrate by parts: $$\int t\,e^{-t}\,dt = -(t+1)\,e^{-t} + K,$$ so

$$-\int t\,e^{-t}\,dt = (t+1)\,e^{-t} + K.$$

Returning to $$x$$ (since $$t=\tfrac{1}{x}$$):
$$\int \frac{e^{-1/x}}{x^{3}}\,dx = \Bigl(\frac{1}{x}+1\Bigr)\,e^{-1/x} + K.$$

Absorb the constant $$K$$ into the general constant $$C$$. Hence from $$(2)$$

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

Multiply by $$e^{1/x}$$:
$$y = 1 + \frac{1}{x} + C\,e^{1/x} \quad -(3)$$

Use the initial condition
Given $$y\!\left(\tfrac{1}{2}\right)=3-e$$. Substitute $$x=\tfrac{1}{2}$$ in $$(3)$$:

$$3-e = 1 + \frac{1}{\tfrac{1}{2}} + C\,e^{\,1/(\tfrac{1}{2})} = 1 + 2 + C\,e^{2} = 3 + C\,e^{2}.$$

Solve for $$C$$:
$$3-e - 3 = C\,e^{2} \;\;\Longrightarrow\;\; -e = C\,e^{2} \;\;\Longrightarrow\;\; C = -\frac{e}{e^{2}} = -\frac{1}{e}.$$

Evaluate $$y(1)$$
Put $$x=1$$ and $$C=-\dfrac{1}{e}$$ into $$(3)$$:

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

Therefore $$y(1)=1$$. The correct choice is Option A.