I'm in the process of refreshing my mind with solving some 'basic' ODE's, I'm relatively proficient with mathematica, however being able to do some of these by hand would be preferable, as I don't always have MMA available to me.

Using an example of 

$$\frac{\partial \left( x \frac{\partial y}{\partial x}\right)}{\partial x}=0$$

The solution is: $$\left\{\left\{y(x)\to c\_1 \ln (x)+c\_2\right\}\right\}$$

I know this solely because MMA told me so.

One possible solution would be to integrate twice for x, 

$$\int \frac{\partial \left(x \frac{\partial y(x)}{\partial x}\right)}{\partial x} \, dx=\int 0 \, dx$$

then

$$\int y'(x) \, dx=\int \frac{c\_1}{x} \, dx$$

with my solution being 

$$y(x)=D[1]+c\_1 \log (x)$$

Great, 

Though if I remember correctly, I can use Z substitution, or D'lambert (though, I can't remember the method, nor with function would be appropriate to replace with)

Trying with Z substitution: 

$$\frac{\partial \left( x \frac{\partial y}{\partial x}\right)}{\partial x}=0$$

I differentiate to get:

$$x y''(x)+y'(x)=0$$

Apply Z sub with z = y' and z' = y''

$$z'=\frac{-z}{x}$$

At this point, I can seperate and intergrate giving me:

$$\frac{x^2}{2}=\frac{z^2}{2}$$

And I know at this point, I am not going to get the correct solution.

So My question is, how do I solve this particular ODE via Substitution or D'lambart

Thanks for the help!