Taking the second derivative of a parametric curve [duplicate]

Consider

$$ \begin{align*} \frac{\text{d}^2y}{\text{d}x^2}&=\frac{\text{d}}{\text{d}x}\left(\frac{\text{d}y}{\text{d}x}\right)\\ &=\frac{\text{d}}{\text{d}t}\left(\frac{\text{d}y}{\text{d}x}\right).\frac{\text{d}t}{\text{d}x}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{1}{\frac{dx}{dt}}\\ \end{align*}$$
where the last equality is as a result of applying the chain rule.

Their justification is that you can use the same process for $\frac{dy}{dx}$ as for $Y$ since you can now consider $Y_2 = g_2(t) = \frac{dy}{dx}(t)$, that is, you once again have a parametric equation in terms of the parameter $t$, and the parametric equation for $x$ stays the same.

I've come up with a 'simple enough' proof for the parametric derivative. Begin by consider a curve parameterized in the way :$<x(t), y(t) >$, then I will rewrite derivative operator with $x$ :

$$ \frac{dy(\theta)}{dx} = \frac{d \theta}{dx} \frac{d y}{d \theta}$$ The above equality holds by the chain rule, derivative of inner function times derivative of outer function, now I will remove the $y$:

$$ \frac{d}{dx} = \frac{d \theta}{dx} \frac{d}{ d \theta} \tag{1}$$

Now, the above can be thought of as an operator which eats a function and maps to another function, for example if we multiply both side on right by $y$ , then it gives the $ \frac{dy}{dx}$ with the parametric derivative of $y$.

And, I will introduce one more identity:

$$ \theta( x( \theta) ) = \theta \tag{2}$$

The above can be thought of in the following way: The function $x(\theta)$ maps from $ \theta \to x$ , the function $ \theta(x)$ maps from $ x \to \theta$, simply the inverse mapping and hence above equality holds true for at least a small interval of where we can say that $ \theta $ and $x$ are bijective functions of each other. Taking the derivate of $(2)$ with $ \theta$,:

$$ \frac{ d \theta}{dx} \frac{dx}{d \theta} = 1$$

Or,

$$ \frac{d\theta}{dx} = \frac{1}{ \frac{ d x}{d \theta}} \tag{2}$$

Hence, we can rewrite (1) as:

$$ \frac{d}{dx} = \frac{ \frac{d }{d \theta} }{\frac{dx}{d \theta}} \tag{3}$$

Now, multiply $y$ on right side of (3): $$ \frac{dy}{dx} = \frac{ \frac{dy }{d \theta} }{\frac{dx}{d \theta}} \tag{4}$$

This is the first parameteric derivative. Now, use (3) on equation (4):

$$ \frac{d^2 y}{dx^2} = \frac{1}{ \frac{dx}{d \theta}} \left[ \frac{ \frac{d^2 y}{d \theta^2} \frac{dx}{d \theta} - \frac{dy}{d \theta} \frac{dx}{ d \theta}}{ ( \frac{dx}{d \theta})^2} \right]= \left[ \frac{ \frac{d^2 y}{d \theta^2} \frac{dx}{d \theta} - \frac{dy}{d \theta} \frac{dx}{ d \theta}}{ ( \frac{dx}{d \theta})^3} \right]$$

Which is what we sought to prove.