Le us consider $\gamma=(x,y)\colon I\to\mathbb R^2$ a parametric curve. Using the relations
- $\mathrm ds=R\mathrm d\theta$,
- $\tan\theta=\frac{\mathrm dy}{\mathrm dx}$,
- $\mathrm ds=\sqrt{\mathrm dx^2+\mathrm dy^2}$,
I managed to prove that the radius of curvature $R$ can be expressed as $$\frac{\left(1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2\right)^{\frac32}}{\frac{\mathrm d^2y}{\mathrm dx^2}}$$
However I also know that: $$R=\frac{(x'^2+y'^2)^{\frac32}}{x'y''-y'x''}$$ with $x'=\frac{\mathrm dx}{\mathrm dt}$ and $y'=\frac{\mathrm dy}{\mathrm dt}$. How to derive this second formula from the first one?
$\begin{align}&\frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx} = \frac{y'}{x' }\\ \\ \Rightarrow&\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{y'}{x'}\right) = \frac{\frac{dy'}{dx}x'-y'\frac{dx'}{dx}}{x'^2} = \frac{\frac{y''}{x'}x'-y'\frac{x''}{x'}}{x'^2}=\frac{x'y''-y'x''}{x'^3}\end{align}$
You can now substitute these and simplify it.