Consider
$$\frac{\partial V}{\partial S} = \frac{1}{S}\frac{\partial V}{\partial x}$$
To find the second order derivative
$$\frac{\partial^2 V}{\partial S^2}$$
we apply the chain rule such that
$$\frac{\partial^2 V}{\partial S^2} \equiv \frac{\partial}{\partial S}\Big[ \frac{1}{S}\frac{\partial V}{\partial x}\Big] = -\frac{1}{S^2}\frac{\partial V}{\partial x} +\frac{1}{S}\frac{\partial}{\partial S}\Big[\frac{\partial V}{\partial x}\Big] $$
I am struggling to solve the seemingly trivial part
$$\frac{1}{S}\frac{\partial}{\partial S}\Big[\frac{\partial V}{\partial x}\Big] $$
The solution should read...
$$\frac{\partial^2 V}{\partial S^2} = -\frac{1}{S^2}\frac{\partial V}{\partial x} + \frac{1}{S^2}\frac{\partial^2 V}{\partial x^2}$$
\begin{align} \frac{\partial^2 V}{\partial S^2} &= \frac{\partial}{\partial S}\left(\frac{\partial V}{\partial S}\right) \\ &= \frac{\partial}{\partial S}\left(\frac{1}{S}\frac{\partial V}{\partial x}\right) \\ &= -\frac{1}{S^2}\frac{\partial V}{\partial x} + \frac{1}{S} \color{blue}{\frac{\partial^2 V}{\partial S \partial x}} \\ &= -\frac{1}{S^2}\frac{\partial V}{\partial x} + \frac{1}{S} \color{blue}{\frac{\partial}{\partial x}\left(\frac{\partial V}{\partial S} \right)} \\ &= -\frac{1}{S^2}\frac{\partial V}{\partial x} + \frac{1}{S} \color{blue}{\frac{\partial}{\partial x}\left(\frac{1}{S}\frac{\partial V}{\partial x}\right)} \\ &= -\frac{1}{S^2}\frac{\partial V}{\partial x} + \frac{1}{S^2}\frac{\partial^2 V}{\partial x^2} \end{align}