$u=\log(x)$ and $v=\log(y)\implies uv=k.$ $uv=k$ satisfies an algebraic differential equation. $X=(x,-y)$ is a vector field with integral curves $uv=k$ in $u-v$ coordinates. A differential equation is $(xy(x))'=0$ which gives solutions $y(x)=k/x.$
Now $\log(x)\log(y)=k$ satisfies the vector field $Y=(x\log(x),-y\log(y)).$ But I can't find the differential equation in this case.
What needs to be done is to map the solution space of $(xy(x))'=0$ to an exponential-exponential coordinate plane and then reconstruct the differential equation. How do you do this exactly?
How does this differential equation embedded in an exponential-exponential coordinate plane change it's form?
I think it will make things more complicated.
Are there any known mathematical problems that are easier to solve in an exponential-exponential coordinate space? Is there a fundamental reason why they seem to be more difficult?