Given differential equation is

$x^{3}dy+\left(xy-1\right)dx=0$

$\Rightarrow \frac{dy}{dx}=\frac{1-xy}{x^3}$

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

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

This is a linear differential equation.

If $\mathrm{IF}=e^{\int \frac{1}{x^2}dx}=e^{-\frac{1}{x}}$

Solution of a given linear differential equation is

$y{e}^{-\frac{1}{x}}=\int e^{-\frac{1}{x}}\left(\frac{1}{x^3}\right)dx$

Put $-\frac{1}{x}=t\Rightarrow \frac{dx}{x^2}=dt$

$\Rightarrow y{e}^{-\frac{1}{x}}=-\int e^{-t}tdt$

$\Rightarrow y{e}^{-\frac{1}{x}}=t{e}^{-t}+e^{-t}+C$

$\Rightarrow y=t+1+C{e}^{\frac{1}{x}}$

$\Rightarrow y=\frac{1}{x}+1+ C{e}^{\frac{1}{x}}$

Put $x=\frac{1}{2}$, then 

$3-e=2+1+C{e}^2$

$C=-\frac{1}{e}$

So,

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

So,

$y\left(1\right)=1+1-1=1$