Answer
An integrating factor is $ \mu (y)=\frac{e^{2y}}{y} $
The differential equation
$$
y d x+\left(2 x y-e^{-2 y}\right) d y=0
$$
is not exact but becomes exact when multiplied by the integrating factor $ \mu (y)=\frac{e^{2y}}{y} $ and its solutions are given implicitly by
$$
x e^{2y} -\ln |y| = c
$$
where $ c $ is an arbitrary constant
Work Step by Step
$$
y d x+\left(2 x y-e^{-2 y}\right) d y=0
\quad \quad (i)
$$
Comparing this Equation with the differential form:
$$
M(x,y) d x+N(x,y) d y=0
$$
we observe that
$$
\begin{aligned}M(x,y) &=y
\\ N(x,y) &=\left(2 x y-e^{-2 y}\right) \end{aligned}
$$
By calculating $M_{y}$ and $N_{x}$ , we find that
$$
\begin{aligned} M_{y}(x, y) &= 1
\\ N_{x}(x, y) & = 2y \end{aligned}
$$
we find that
$$
\frac{N_{x}(x, y)-M_{y}(x, y)}{M(x, y)}=\frac{2y -1 }{y }=2-\frac{1}{y}.
$$
Thus there is an integrating factor $μ $ that is a function of $y$ only, and it satisfies the differential equation
$$
\frac{d \mu}{d y}=\left(2-\frac{1}{y}\right) \mu .
$$
Hence
$$
\frac{d \mu}{ \mu}=\left(2-\frac{1}{y}\right) dy
$$
Integrating, we obtain
$$
\ln \mu=2y- \ln y= \ln e^{2y}- \ln y =\ln \left( \frac{e^{2y}}{y}\right)
$$
take the exponentiation we have
$$
\mu (y)=\frac{e^{2y}}{y}
$$
Multiplying Eq. (i) by this integrating factor, we obtain
$$
e^{2y} d x+\left(2 x e^{2 y} - \frac{1}{y}\right) d y=0
$$
The latter equation is exact, since
$$
M_{y}(x, y)=2 e^{2 y} =N_{x}(x, y).
$$
Thus there is a $\psi (x, y)$ such that
$$
\begin{aligned} \psi_{x}(x, y) &=M(x,y) =e^{2y}
\\
\psi_{y}(x, y) &=N(x,y) =\left(2 x e^{2 y} - \frac{1}{y}\right)
\end{aligned} \quad \quad (ii)
$$
Integrating the first of these equations with respect to $x$ , we obtain
$$
\begin{split}
\psi(x, y) & =\int { e^{2y} } dx \\
& = x e^{2y} +h(y)
\quad \quad \quad\quad\quad (iii)
\end{split}
$$
where $h(y)$ is an arbitrary function of $y$ only. To try to satisfy the second of Eqs. (ii), we compute $\psi_{y}(x, y) $ from Eq. (iii) and set it equal to $N$, obtaining
$$
\psi_{y}(x, y)= 2x e^{2y} +h^{\prime}(y)= \left(2 x e^{2 y} - \frac{1}{y}\right)
$$
Thus $h^{\prime}(y)=- \frac{1}{y} $ and $h(y)=-\ln y $ , the constant of integration can be omitted since any solution of the preceding differential equation is satisfactory.
Now substituting for $h(y)$ in Eq. (iii) gives
$$
\psi(x, y)= x e^{2y} -\ln |y|
$$
Hence solutions of Eq. (i) are given implicitly by
$$
\psi(x, y) = x e^{2y} -\ln |y| =c
$$
where $ c $ is an arbitrary constant