Let $f:{\mathbb{R}}^{n}\to \mathbb{R}$ be a differentiable function. Remember that the gradient of function f is defined in the following way:

$\mathrm{\nabla}f=(\frac{df}{{dx}_{1}},\frac{df}{{dx}_{2}},\dots ,\frac{df}{{dx}_{n}})$

To calculate $\frac{df}{dx}$ imagine that y=C for some constant $C\in \mathbb{R}$ and take the derivative with respect to x. To obtain $\frac{df}{dy}$

$\frac{df}{dx}(x,y)=\frac{d}{dx}\frac{1}{\sqrt{{x}^{2}+{C}^{2}}}=\frac{-1}{2}\frac{2x}{{({x}^{2}+{C}^{2})}^{\frac{3}{2}}}$

$=-\frac{x}{{({x}^{2}+{C}^{2})}^{\frac{3}{2}}}=-\frac{x}{{({x}^{2}+{y}^{2})}^{\frac{3}{2}}}$

Employing the analogous tehnique or observing the symmetry yields:

$\frac{df}{dy}(x,y)=-\frac{y}{{({x}^{2}+{y}^{2})}^{\frac{3}{2}}}$ Which gives

$\mathrm{\nabla}sf(x,y)=(-\frac{x}{{({x}^{2}+{y}^{2})}^{\frac{3}{2}}},-\frac{y}{{({x}^{2}+{y}^{2})}^{\frac{3}{2}}}$