Condition for the directional derivative of $f$ along the circle’s tangent to vanish

Parametrize the unit circle by $\gamma(t)=(\cos t,\sin t)$. The unit tangent at a point $P=(x,y)$ is $E_t=\gamma'(t)=(-\sin t,\cos t)=(-y,x)$. For any differentiable $f:\mathbb{R}^2\to\mathbb{R}$ the directional derivative of $f$ in direction $E_t$ at $P$ is \[ d f_P(E_t)=f_x(P)(-y)+f_y(P)\,x. \] Thus $d f_P(E_t)=0$ exactly when \[ -\,y\,f_x(x,y)+x\,f_y(x,y)=0,\qquad\text{i.e. }\; x f_y(x,y)=y f_x(x,y). \] Geometrically this means the gradient $\nabla f(P)$ is parallel to the radius vector $(x,y)$ (equivalently $f$ is locally constant along the circle at $P$). If a particular $f$ is given, substitute its partial derivatives into this relation and solve for the points $(x,y)$ on the circle.