Understanding this partial derivative problem 
Problem:
Considering $x$ and $y$ as independent
   variables, find $\dfrac{\partial
 r}{\partial x}, \dfrac{\partial
 r}{\partial y}, \dfrac{\partial
 \theta}{\partial x}, \dfrac{\partial
 \theta}{\partial y}$ when $x = e^{2r}
 \cos \theta, y = e^{3r} \sin \theta$.
Solution:
First differentiate the given
   relations with respect to $x$:
$1 = 2e^{2r} \cos \theta
 \dfrac{\partial r}{\partial x} - e^{2r}
 \sin \theta \dfrac{\partial
 \theta}{\partial x}$ and $0 =
 3e^{3r}\sin \theta \dfrac{\partial
 r}{\partial x} + e^{3r} \cos \theta
 \dfrac{\partial \theta}{\partial x}$.
Then solve simultaneously to obtain
   $\dfrac{\partial r}{\partial x} =
 \dfrac{\cos \theta}{e^{2r}(2+\sin^{2}
 \theta)}$ and $\dfrac{\partial
 \theta}{\partial x} = - \dfrac{3 \sin
 \theta}{e^{2r}(2+sin^2 \theta)}$

Question:
(1)
So first of all, why does differentiating with respect to $x$ result in $\dfrac{\partial r}{\partial x}$ and $\dfrac{\partial \theta}{\partial x}$ (by the way what are these called?)? Is this because the problem says "$x$ and $y$ as independent variables" ? My initial reaction was to do $r$ and $\theta$ separately while regarding all other variables as constants... This is implicit (partial?) differentiation, right?
How should I understand what is being done here? A lot of times it seems that things like this turn out to really just be a mapping. Can I think of this that way as well? I couldn't even say what the domain and codomain would be...
Whenever I perform this operation, do I just take partial derivatives of both sides treating all independent variables (other than the variable with respect to which I am differentiating) as constants and all non-independent (dependent?) variables as variables that need to be differentiated and will have that partial derivative symbol? Once I accept that I can see how they got the first implicit partial differentiation (if that's what it is called).
(2)
What do they mean by "solve simultaneously" ? I tried to solve for each (again I don't know what they're called yet) "partial differential" resulting in: $$\dfrac{\partial \theta}{\partial x} = -\dfrac{\sin \theta \frac{\partial r}{\partial x}}{\cos \theta}$$ and $$\dfrac{\partial r}{\partial x} = \dfrac{1+e^{2r}\sin \theta \frac{\partial \theta}{\partial x}}{2e^{2r}\cos \theta}$$
But I couldn't get the same answer... I tried to substitute this further, but looking at that answer, I was sure I was missing something... Could somebody please show me what to do?
Thank you in advance for any help!    
 A: *

*That's just the Chain Rule (with partials because you know there are 'really' two variables around). Think of $r$ as a function $r(x,y)$, and $\theta$ as a function $\theta(x,y)$, and you are just doing implicit differentiation/chain rule. The fact that "$x$ and $y$ are independent variables" tells you that $\frac{dy}{dx} = 0$, so that's why you get that differentiating the relation $y=e^{3r}\cos\theta$ with respect to $x$ gives you $0$ on the left hand side.


*You have two expressions, both involving the unknown functions $\frac{\partial r}{\partial x}$, $\frac{\partial \theta}{\partial x}$. Think of those two as the "unknowns" in a $2\times 2$ system, and solve the system (like you would a regular system of two equations in two unknowns) to get expressions for each involving only $r$s and $\theta$s, and no partials, $x$s, or $y$s.
E.g., take the second equation and multiply through by $2\cos\theta$ to get
$$0 = 6e^{3r}\sin\theta\cos\theta\frac{\partial r}{\partial x} + 2e^{3r}\cos^2\theta\frac{\partial\theta}{\partial x}.$$
Then take the first equation and multiply through by $-3e^r\sin\theta$ to get
$$-3e^r\sin\theta = -6e^{3r}\sin\theta\cos\theta\frac{\partial r}{\partial x} +3e^{3r}\sin^2\theta \frac{\partial\theta}{\partial x}.$$
Adding both equations eliminates $\frac{\partial r}{\partial x}$, so now you can solve for $\frac{\partial\theta}{\partial x}$. Etc.
Then you do the same thing but with partials with respect to $y$, using the fact that $x$ and $y$ are independent to get that $\frac{dx}{dy}=0$.
A: 
(1) So first of all, why does differentiating with respect to x result in $\frac{\partial r}{\partial x}, \frac{\partial \theta}{\partial x}$

You're using multidimension chain rule. You want to take a derivative of a function $f(r,\theta)$ with respect to $x$, but both $r$ and $\theta$ depend on $x$. So $\dfrac{d}{dx}x = \dfrac{d}{dx}\left(e^{2r}
 \cos \theta\right) = $ by chain rule $ \dfrac{\partial}{\partial r}\left(e^{2r}\cos\theta\right)\dfrac{\partial r}{\partial x} + \dfrac{\partial}{\partial\theta}\left(e^{2r}\cos\theta\right)\dfrac{\partial\theta}{\partial x}$.
Similalry $\dfrac{\partial{f(r,\theta)}}{\partial x} = \dfrac{\partial{f(r,\theta)}}{\partial r}\dfrac{\partial r}{\partial x} +\dfrac{\partial{f(r,\theta)}}{\partial \theta}\dfrac{\partial \theta}{\partial x}$
The same is true for $y$. I hope this helps... I'm not sure if this is your question, since you used the chain rule fine. Let me know and I'll revise...
