I am trying to prove the parts of each component of the inverse matrix in the attached image. I have tried using differentials and then solving for the other components. (I'd like to solve it this way). Trying to solve for example, $\frac{d\theta}{dx}$ (in the bottom left of the inverse matrix [attached below]) $$x = r cos(\theta)$$ –> $$dx = cos(\theta)dr - rsin(\theta)d\theta$$Then observing that we are holding $r = constant$, thus $dr = 0$. I get that $\frac{d\theta}{dx} = \frac{- 1}{r sin(\theta)}$, which is close. I put this in a partial calculator and made $\theta$ a function of x and r, $\theta = cos^{-1}\left(\frac{x}{r}\right) = \cos^{-1}\left(\frac{x}{\sqrt{x^2 +y^2}}\right)$. Taking the $\frac{\partial \theta}{\partial x}$ I get the right answer due to r being a function of x and y. If I use the $cos^{-1}\left(\frac{x}{r}\right)$ and take the partial I get what I stated above ($\frac{d\theta}{dx} = \frac{- 1}{r sin(\theta)}$). Also I tried to replace dr in $dx = cos(\theta)dr - rsin(\theta)d\theta$ by using $r^2=x^2+y^2$ by replacing dr with $rdr = xdx + ydy$ where I assumed dy to be constant. Which yielded me the wrong answer. I'd like to improve my logical thinking so any advice on what I did would be great as well. Thank you!
Summary: I am trying to prove using differentials (not partials) that $\frac{d\theta}{dx} = \frac{-sin(\theta)}{r}$
The issue is that you cannot just write $\frac{d\theta}{dx}$. In thermodynamics, there is a notation which really is useful and important. They write partial derivatives with a subscript to indicate what variable(s) are left fixed. So, for example, if we have $z=f(x,y)$ and we want to find the derivative of $f$ with respect to $x$, fixing $y$, we write $$\left(\frac{\partial f}{\partial x}\right)_y \quad\text{or}\quad \left(\frac{\partial z}{\partial x}\right)_y.$$ This is important because we may have lots of variables flying around and it's important to know what variable(s) are fixed.
In your example, we can think of $(x,y)$ as functions of $(r,\theta)$. Then if we write $\partial x/\partial\theta$, this ordinarily signifies $\left(\frac{\partial x}{\partial\theta}\right)_r$. When you fix $r$, then it becomes true (because we're essentially doing one-dimensional calculus) that $$\left(\frac{\partial\theta}{\partial x}\right)_r = \frac 1{\left(\frac{\partial x}{\partial\theta}\right)_r}.$$ However, you're muddling things by instead trying to compute $\left(\frac{\partial\theta}{\partial x}\right)_y$, and these are two totally different beasts. You really must be careful about keeping track of the independent variables. If you change those, more chain rule comes in.
Just to reiterate, you are trying to compare \begin{align*} \left(\frac{\partial\theta}{\partial x}\right)_r &= -\frac1{r\sin\theta} = -\frac1y \quad\text{and} \\ \left(\frac{\partial\theta}{\partial x}\right)_y &= -\frac{y}{x^2+y^2} = -\frac{\sin\theta}r. \end{align*}
By the way, be warned. In general, we do not have $\frac{\partial x}{\partial\theta} = \frac1{\frac{\partial\theta}{\partial x}}$. Indeed, since $x=r\cos\theta$, we have $\partial x/\partial\theta = -r\sin\theta$ (which is $-y$). On the other hand, since $\theta =\arctan(y/x)$ (at least for $-\pi/2<\theta<\pi/2$), we have $\partial\theta/\partial x = -\frac y{x^2+y^2}$, which is very different from $-y$. This is your $-\sin\theta/r$, of course. The correct relation comes from the complete derivative matrices (called Jacobians), which are inverse $2\times 2$ matrices.
You can do this all correctly with differentials (differential forms, in fact), but you must still keep track of who the independent variables are. And you really must stop writing things like $d\theta/dx$ unless $\theta$ really is a function just of one variable $x$. To get your first formula, you would have to write $d\theta$ in terms of just $dx$ and $dr$; to get the second you would have to write $d\theta$ in terms of the usual $dx$ and $dy$. It's just a question of what the independent variables are.