Understanding this partial derivative problem

1. 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.

2. 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 $\frac{dx}{dy}=0$.

(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...