As $r$ is given, I calculate some derivatives using Wolfram Alpha:
$$r=\sqrt{(x-a)^2+(y-b)^2}$$
$$\frac{\partial r}{\partial x}=\frac{x-a}{r}$$ $$\frac{\partial r}{\partial a}=-\frac{\partial r}{\partial x}$$ $$\frac{\partial^2 r}{\partial x^2}=\frac{1}{r}-\frac{(x-a)^2}{r^3}$$
This equality
$$\frac{\partial^2 r}{\partial a^2}=-\frac{\partial^2 r}{\partial x^2}$$
I try to prove it myself
$$\frac{\partial^2 r}{\partial a^2}=\frac{\partial}{\partial a}\frac{\partial r}{\partial a}=\frac{\partial }{\partial a}(-\frac{\partial r}{\partial x})=-\frac{\partial }{\partial a}\frac{\partial r}{\partial x}=-\frac{\partial }{\partial x}\frac{\partial r}{\partial a}=-\frac{\partial }{\partial x}(-\frac{\partial r}{\partial x})=\frac{\partial^2 r}{\partial x^2}$$
Obviously this proof is incorrect. Where I made a mistake? Can't see
This is because what you were trying to prove is wrong. It is actually true that $$\frac{\partial^2 r}{\partial a^2}=\frac{\partial^2 r}{\partial x^2},$$ which is the result of your (correct) proof.