$y' = \exp(-\frac{y}{x}) + \frac{y}{x}$ Could you help me to solve this differential equation:
$$y' = \exp\left(-\frac{y}{x}\right) + \frac{y}{x}$$

Following the hint of user71352 we substitute
$$y(x) = x\cdot v(x), \qquad y'(x) = v(x) + x\cdot v'(x)$$
back  and we get
\begin{align*}
v(x) + xv'(x) &=e^{-v(x)} + v(x) \\
xv'(x) &= e^{-v(x)} \\
v'(x) &= \dfrac{e^{-v(x)}}{x} \\
e^{v(x)}v'(x)&= \dfrac{1}{x} \\
\int e^{v(x)}v'(x) \, dx&= \int\dfrac{1}{x}\, dx \\
e^{v(x)} &= \log(x) + C \\
v(x) &= \log(\log(x)+C) \implies y(x) = x\log(\log(x)+C)
\end{align*}
where $C$ is an arbitrary constant
