I am having trouble thinking about $\text{dy}$ and $\text{dx}$, $\text{d}^2\text{y}$ and $\text{dx}^2$ as differentials. I get that you can write $\text{dy}=f'(x)\text{dx}$ and how to derive it but then we get to writing $\text{d}^2\text{y} = \text{d(dy)} = \text{d}(f'(x)\text{dx}) = f''(x)\text{(dx)}^2$ and I am lost on the last 2 steps.
Can someone explain how $\text{d}(f'(x)\text{dx}) = f''(x)\text{(dx)}^2$, particularly how $\text{d}(f'(x)\text{dx})=f''(x)(\text{dx}\cdot\text{dx})$. I don't know where the extra $\text{dx}$ comes from. Is it from $\text{d(dx)}$, is a chain rule involved with $\text{d}(f'(x)\text{dx})$ where you treat $f'(x)$ and $\text{dx}$ as 2 functions being affected by "$\text{d}$".
I think my confusion stems from not knowing what the "$\text{d}$" really means. Sometimes I think its an operator of sorts so like $\text{d}(u v)$ means to differentiate (kind of) to get $u \text{dv} + v \text{du}$ but then I wonder what $\text{d}(x^2\text{dx})$ would be and I don't know how to treat the combination of $\text{d}$ and $\text{dx}$...
What is really going on is $$ y = f(x) \Rightarrow \frac{d^2}{dx^2} y = \frac{d^2}{dx^2} f(x) = \frac{d}{dx} \frac{d}{dx} f(x) = \frac{d}{dx} f'(x) = f''(x) $$ where $d/dx$ is the operation of taking the derivative regarding $x$ and $d^2/dx^2$ means applying it two times.
Then multiply the above equation chain "formally" by $dx^2$ and compare with your results. I think your $d$ should be a $d^2$ at some points.