Curl of $\frac{\hat r}{r^2}$ using two different coordinates

Question

I am learning vector calculus. Here I wanted to take out the $\nabla\times(\frac{\hat r}{r^2})$ , So in spherical coordinates it is easy to take out. It is zero. but while doing in Cartesian coordinates $\begin{bmatrix} \hat x & \hat y & \hat z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ \frac{1}{(x^2+y^2+z^2)} & \frac{1}{(x^2+y^2+z^2)} & \frac{1}{(x^2+y^2+z^2)} \\ \end{bmatrix} $

This on solving isn't coming to be zero. Why?

Answer 1

$\frac{\hat{r}}{r^2} = \frac{(x,y,z)}{(x^2+y^2+z^2)^\frac{3}{2}} \neq \frac{(1,1,1)}{x^2+y^2+z^2}$

Thanks to Ninad Munshi.