Let f(x,y)=1x2+y2. Compute ∇f(x,y)

Let ${\displaystyle f:{\mathbb{R}}^n\to \mathbb{R}}$ be a differentiable function. Remember that the gradient of function f is defined in the following way:
${\displaystyle \mathrm{\nabla }f=(\frac{df}{{dx}_1},\frac{df}{{dx}_2},\dots ,\frac{df}{{dx}_n})}$
To calculate ${\displaystyle \frac{df}{dx}}$ imagine that y=C for some constant ${\displaystyle C\in \mathbb{R}}$ and take the derivative with respect to x. To obtain ${\displaystyle \frac{df}{dy}}$
${\displaystyle \frac{df}{dx}(x,y)=\frac{d}{dx}\frac{1}{\sqrt{x^2+C^2}}=\frac{-1}{2}\frac{2x}{{(x^2+C^2)}^{\frac{3}{2}}}}$
${\displaystyle =-\frac{x}{{(x^2+C^2)}^{\frac{3}{2}}}=-\frac{x}{{(x^2+y^2)}^{\frac{3}{2}}}}$
Employing the analogous tehnique or observing the symmetry yields:
${\displaystyle \frac{df}{dy}(x,y)=-\frac{y}{{(x^2+y^2)}^{\frac{3}{2}}}}$ Which gives
${\displaystyle \mathrm{\nabla }sf(x,y)=(-\frac{x}{{(x^2+y^2)}^{\frac{3}{2}}},-\frac{y}{{(x^2+y^2)}^{\frac{3}{2}}}}$