Solve $(x+y \log y)y\, \mathrm{d}x=(y+x \log x)x \,\mathrm{d}y$ Solve the differential equation $$(x+y \log y)y \,\mathrm{d}x=(y+x \log x)x \,\mathrm{d}y$$
My try: put $x=e^t$ and $y =e^w$: we get
$$(e^t+w e^w)e^w e^t \,\mathrm{d}t=(e^w+te^t)e^we^t\,\mathrm{d}w\implies
(e^t+we^w)\,\mathrm{d}t=(e^w+te^t)\,\mathrm{d}w$$
Any clue here?
 A: We have the following differential equation:

$$(x+y \log y)y \,\mathrm{d}x=(y+x \log x)x \,\mathrm{d}y$$

Rearranging the terms, we get:
$$\implies xy\mathrm{d}x+y^{2}\log(y)\mathrm{d}x=xy\mathrm{d}y+x^{2}\log(x)\mathrm{d}y$$
$$\implies-x^{2}\log(x)\mathrm{d}y+xy\mathrm{d}x=-y^{2}\log(y)\mathrm{d}x+xy\mathrm{d}y$$
$$\implies x(-x\log(x)\mathrm{d}y+y\mathrm{d}x)=y(-y\log(y)\mathrm{d}x+x\mathrm{d}y)$$
Dividing both sides by the term $x^{2}y^{2}$,  we get:
$$\implies \frac{(-x\log(x)\mathrm{d}y+y\mathrm{d}x)}{xy^2}=\frac{(-y\log(y)\mathrm{d}x+x\mathrm{d}y)}{x^{2}y}$$
$$\implies \left(-\frac{1}{y^2}\mathrm{d}y\right)\log(x)+\left(\frac{1}{x}\mathrm{d}x\right)\frac{1}{y}=\left(-\frac{1}{x^2}\mathrm{d}x\right)\log(y)+\left(\frac{1}{y}\mathrm{d}y\right)\frac{1}{x}$$
$$\implies \mathrm{d}\left(\frac{1}{y}\right)\log(x)+\mathrm{d}\left(\log(x)\right)\frac{1}{y}=\mathrm{d}\left(\frac{1}{x}\right)\log(y)+\mathrm{d}\left(\log(y)\right)\frac{1}{x}$$
$$\implies \mathrm{d}\left(\frac{1}{y}\log(x)\right)=\mathrm{d}\left(\frac{1}{x}\log(y)\right)$$
Integrating both sides, we get:
$$\implies \int \mathrm{d}\left(\frac{1}{y}\log(x)\right)=\int \mathrm{d}\left(\frac{1}{x}\log(y)\right)$$
We get the solution as: $$\color{darkblue}{\frac{1}{y}\log(x)=\frac{1}{x}\log(y)+C}$$
Where $C$ is the integration constant. This is the implicit solution.
Edit: As shown by @Claude Leibovici, the solution can be converted to an explicit one using Lambert $W$ function. The solution is as follows:
Rearranging the above equation, we get:
$$\implies \frac{x}{y}\log(x)=\log(y)+Cx$$
Taking exponential of both sides, we have:
$$\implies \exp\left({\frac{x}{y}\log(x)}\right)=\exp\left(\log(y)\right)\exp\left(Cx\right)$$
$$\implies \exp\left({\frac{x}{y}\log(x)}\right)=y\exp\left(Cx\right)$$
Multiplying both sides by $\frac{x}{y}\log(x)$, we have:
$$\implies \frac{x}{y}\log(x)\exp\left({\frac{x}{y}\log(x)}\right)=x\exp\left(Cx\right)\log(x)$$
If we consider $\frac{x}{y}\log(x)$ as $u$, the above equation reduces to $ue^u=K$ and can be represented by Lambert $W$ function as $u=W(K)$, so we have:
$$\implies \frac{x}{y}\log(x)=W\left(x\exp\left(Cx\right)\log(x)\right)$$
$$\implies \color{darkgreen}{y=\frac{x \log (x)}{W\left(x e^{Cx} \log (x)\right)}}$$
where $W(.)$ is Lambert function. This is the explicit solution.
