Let
$$r=\sqrt{(x-x_i)^2+(y-y_i)^2}$$
$$\frac{dr}{dx}=\frac{x-x_i}{r}$$
$U(r)$ is potential energy. It is known, that
$$\psi=\frac{dU}{dr}=-(k+\frac{1}{r})U$$ $$\frac{\partial U}{\partial x}=\frac{dU}{dr}\cdot\frac{\partial r}{\partial x}=\psi\cdot\frac{\partial r}{\partial x}=-\frac{x-x_i}{r}(k+\frac{1}{r})U$$
$k$ is some constant.
I need to compute $\frac{\partial ^2U}{\partial x^2}$ and I fail. And I do not see where I did a mistake. Please check this out
My attempt:
$$\frac{\partial ^2U}{\partial x^2}=\frac{\partial }{\partial x}(\psi\frac{\partial r}{\partial x})=\frac{\partial \psi}{\partial x}\cdot\frac{\partial r}{\partial x}+\psi\frac{\partial ^2r}{\partial x^2}=\frac{d\psi}{dr}(\frac{\partial r}{\partial x})^2+\psi\frac{\partial ^2r}{\partial x^2}$$
I find
$$\frac{d\psi}{dr}=(k^2+\frac{2k}{r}+\frac{2}{r^2})U$$
When substituting everything back I get
$$\frac{d\psi}{dr}(\frac{\partial r}{\partial x})^2+\psi\frac{\partial ^2r}{\partial x^2}=(k^2+\frac{2k}{r}+\frac{2}{r^2})U\cdot\frac{(x-x_i)^2}{r^2}-(k+\frac{1}{r})U\cdot\frac{1}{r}$$ $$\frac{\partial ^2U}{\partial x^2}=\frac{1}{r}U(\frac{(x-x_i)^2}{r}(k^2+\frac{2k}{r}+\frac{2}{r^2})-(k+\frac{1}{r}))$$
Which is wrong answer. The right answer is
$$\frac{\partial ^2U}{\partial x^2}=\frac{1}{r}U(\frac{(x-x_i)^2}{r}(k^2+\frac{3k}{r}+\frac{3}{r^2})-(k+\frac{1}{r}))$$
Where did I do a mistake?
You used $$\frac {d^2 r}{dx^2}=\frac 1r$$ It's not true. After you took the first derivative, you ignored that $r=r(x)$
$$\frac {d^2 r}{dx^2}=\frac 1r-\frac{(x-x_i)^2}{r^3}$$