A useful tip: it’s often easier to do as much as you can before you substitute some possibly messy specific function into a general formula.
Your first differences are $y(x_0+\Delta x)-y(x_0)$ and $y(x_0+2\Delta x)-y(x_0+\Delta x)$, so your second difference is
$\Big(y(x_0+2\Delta x)-y(x_0+\Delta x)\Big)-\Big(y(x_0+\Delta x)-y(x_0)\Big)\;.$
This simplifies to
$y(x_0+2\Delta x)-2y(x_0+\Delta x)+y(x_0)\;,$
which becomes $\frac1{x_0+2\Delta x}-\frac2{x_0+\Delta x}+\frac1{x_0}$ when we plug in the actual function. Combining this over the least common denominator, we get $\frac{x_0(x_0+\Delta x)-2x_0(x_0+2\Delta x)+(x_0+\Delta x)(x_0+2\Delta x)}{x_0(x_0+\Delta x)(x_0+2\Delta x)}\;.$ The numerator expands to
$x_0^2+x_0\Delta x-2x_0^2-4x_0\Delta x+x_0^2+3x_0\Delta x+2\Delta x^2\;,$
and everything cancels out except the $2\Delta x^2$, so your second difference is just $\frac{2\Delta x^2}{x_0(x_0+\Delta x)(x_0+2\Delta x)} ;,$ so $\frac{\Delta^2y}{\Delta x^2}=\frac2{x_0(x_0+\Delta x)(x_0+2\Delta x)}\;,$ and from here you should be home free.