Question: $$x\frac{dy}{dx}=y(1+\ln y-\ln x)$$
I realize that it can be rearranged to make it clear that it's a homogeneous first order differential:
$$\frac{dy}{dx}=\frac yx(1+\ln \frac yx)$$
Using the substitution $v=\frac yx$ and using the chain rule I obtain:
$$x\frac{dv}{dx}+v=v(1+\ln v)$$
$$x\frac{dv}{dx}=v\ln v$$
The differential equation then becomes:
$$\int\frac{dv}{v\ln v}=\int\frac{dx}{x}$$
Is this correct? How do integrate the integral on the left hand side?

Answer: 
That looks okay. You may integrate the left side by using the change of variable $u=\ln(v)$, $du=\frac1v dv$.
$$x\frac{dy}{dx}=y(1+\ln y-\ln x)$$

$$\frac{1}{y}\frac{dy}{dx}=\frac{1+\ln y-\ln x}{x}.~.~.~.~(1)$$

putting $\ln{y}=v$ we have $$\frac{1}{y}\frac{dy}{dx}=\frac{dv}{dx}$$

now putting it in equation (1), we have $$\frac{dv}{dx}=\frac{1+v-\ln x}{x}$$

$$\frac{dv}{dx}-\frac{v}{x}=\frac{1-\ln x}{x}$$

which is linear differential equation in $v$ and therefore 

integrating factor $=e^{\int -\frac{1}{x}dx} =\frac{1}{x}$

and therefore solution

$$v\frac{1}{x}=\int {\Big(\frac{1-\ln x}{x}\Big)\frac{1}{x} }~dx~+~\ln C$$

The final answer will be $$y=xe^{Cx}$$