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.