vector calculus - show field is conservative

If F = $\displaystyle -2\frac{xi + yj + zk}{(x^2 + y^2 + z^2)^2} $

Show by calculating $\displaystyle \nabla x F $ that vector field F is conservative.

I know that $\displaystyle \nabla x F = Curl F = 0$

And then I do the determinant with i, j and k across the top row, and $\displaystyle \frac{\partial}{\partial x}, \frac{\partial}{\partial y} and \frac{\partial}{\partial z} $ across the second row.

But I'm confused as to what I put along the 3rd row since F is a fraction...

Would i split it up into $\displaystyle \frac{-2x}{(x^2 + y^2 + z^2)^2}, \frac{-2y}{(x^2 + y^2 + z^2)^2} $and$\displaystyle \frac{-2z}{(x^2 + y^2 + z^2)^2}$

Thanks in advance!