Level curves and gradients

Level curves. $f=c$ is the circle $x^2+y^2=c$ of radius $\sqrt c$ (for $c>0$).

Parameterization. $\gamma(t)=(\sqrt c\cos t,\sqrt c\sin t)$; $\gamma'(t)=\sqrt c(-\sin t,\cos t)$ is tangent.

Gradient. $\nabla f=2(x,y)=2\gamma(t)$, radial and orthogonal to $\gamma'(t)$:
$$(2\sqrt c\cos t,2\sqrt c\sin t)\cdot(-\sqrt c\sin t,\sqrt c\cos t)=0.$$

Conclusion. $\nabla f$ is orthogonal to the level curves; it points in the direction of fastest increase of $f$.