given that u = f(y) and y = g(x), could anyone prove that $$ \frac{d^2u}{dx^2} = \frac{du}{dy} \frac{d^2y}{dx^2} + \frac{d^2u}{dy^2}\left(\frac{dy}{dx}\right)^2$$?
I've tried to use the chain rule, but this is all I've got: $$\frac{d^2u}{dx^2} = \frac{d}{dx}\frac{du}{dx} = \frac{d}{dx}\left(\frac{du}{dy}\frac{dy}{dx} \right) = \frac{du}{dy}\frac{d^2y}{dx^2} + \frac{dy}{dx} \frac{d}{dx}\left(\frac{du}{dy}\right)$$
You're almost there. Now, just notice that (by chain rule again) \begin{align} \dfrac{d}{dx} \left( \dfrac{du}{dy}\right) &= \dfrac{d}{dy} \left( \dfrac{du}{dy}\right) \cdot \dfrac{dy}{dx}\\ &= \dfrac{d^2u}{dy^2} \cdot \dfrac{dy}{dx}. \end{align} If you plug this into your expression, you'll get the desired form.
If it makes it clearer, temporarily call $v = \dfrac{du}{dy}$. Then, \begin{align} \dfrac{dv}{dx} &= \dfrac{dv}{dy} \cdot \dfrac{dy}{dx} \\ &= \dfrac{d^2u}{dy^2} \cdot \dfrac{dy}{dx}. \end{align}