I am working on some practice exercises (not homework) on transformations and need some intuition and help.
One of the questions is: $(u,v)=f(x,y)$
where $ \quad u= { e }^{ x }\cos(y), \quad v = { e }^{ x }\sin(y).\quad \quad $
A) It asks to compute the Jacobian $\operatorname{det} Df$ which is ${ e }^{ 2x }$, next it asks to find the formulas for the local inverses of $f$ when they exist. How do I find inverses when there are two variables involved?
B) It also asks to sketch of the lines $x=\text{constant}$ and $y = \text{constant}$ and I believe if I let $x$ to be some constant $c$ then I get $(u,v)$ to be $(c\cos(y),c\sin(y))$ but how do I graph this?
For Part A, you need to find $(x,y)=(x(u,v), y(u,v))$. You might start by writing $u^{2}+v^{2}$ as a function of $x$ and $y$. That should make it easy for you to find $x(u,v)$.
For Part B, you have correctly derived the expression describing the transformation of lines of constant $x$. Think about polar coordinates, and you should see what shape that expression describes in the $(u,v)$-plane. Lines of constant $y$ transform to similarly simple curves in the $(u,v)$-plane.
EDIT TO ADD:
Okay, so you've pretty much gotten through it, I think, with the help of some comments, but here's the remainder of Part A for completeness:
$$u^{2}+v^{2} = e^{2x}\cos^{2}y + e^{2x}\sin^{2}y = e^{2x},$$
and, therefore, $$x = {1\over2}\log(u^2+v^2).$$
And you correctly derived that $$y=\tan^{-1}{v\over u}.$$
Now note that this is, indeed, only a local inversion, because the arctangent is not uniquely defined (you have to specify the particular range of values, for example, $-\pi/2\lt y \lt\pi/2$).
Do you now have a better idea of the curves on the $(u,v)$-plane described by lines of constant $x$ and $y$?