By a suitable substitution or differentiation, reduce…
By a suitable substitution or differentiation, reduce $x(e^y-y')=2$ to linear form and find its general integral.
Explanation of Method
Isolate $y'=e^y-\dfrac2x$, then substitute $u=e^{-y}$ so that $u'=-e^{-y}y'$; this turns the equation into a linear equation for $u(x)$.
Step 1: Isolate $y'$
From $x(e^y-y')=2$: $\;y'=e^y-\dfrac{2}{x}$.
Step 2: Substitute $u=e^{-y}$
Let $u=e^{-y}$, so $u'=-e^{-y}y'=-uy'$, i.e. $y'=-u'/u$. Also $e^y=1/u$. Substituting:
\[-\frac{u'}{u}=\frac1u-\frac2x.\]
Multiply by $-u$:
\[u'=-1+\frac{2u}{x}\quad\Longrightarrow\quad u'-\frac{2}{x}u=-1.\]
Step 3: Solve the linear equation in $u$
The integrating factor is $\mu(x)=e^{-\int \frac2x dx}=x^{-2}$, giving $\dfrac{d}{dx}\!\left(\dfrac{u}{x^2}\right)=-\dfrac{1}{x^2}$, so
\[\frac{u}{x^2}=\frac1x+C\quad\Longrightarrow\quad u=x+Cx^2.\]
Step 4: Back-substitute
Since $u=e^{-y}$:
\[e^{-y}=x+Cx^2.\]
Result
Final answer:
\[\boxed{\,e^{y}\big(x+Cx^2\big)=1,\qquad C\in\mathbb{R}}\]