gradient is perpendicular to level curves

$\nabla f$ always points exactly perpendicular to the level curve passing through that point.

Along a level curve $f(x,y)=k$ , $f$ doesn't change at all by definition, so $\nabla f$ can't have any component pointing along that curve — differentiating $f(x(t),y(t))=k$ , a curve constrained to the level set, gives $\nabla f\cdot\mathbf{r}'(t)=0$ for free from the chain rule. This identifies the tangent line to an implicitly-defined curve directly from $\nabla f$ , with no need to ever isolate $y$ as a function of $x$ .

Examples

On the folium $x^3+y^3=9xy$ at $(2,4)$ , $\nabla F(2,4)=\langle-24,30\rangle$ gives tangent line $-4x+5y=12$ .