Represent the level curve in the position-vector form $\mathbf{R}\=\left[\begin{array}{c}x\\ y\(x\)\end{array}\right]$ so a vector tangent to this curve is then $\mathbf{R}\prime \left(x\right)\=\left[\begin{array}{c}1\\ y\prime \(x\)\end{array}\right]$.

By implicitly differentiating $f\left(x\,y\left(x\right)\right)\equiv c$ to get $f_x\+f_y y\prime \=0$, the derivative $y\prime \left(x\right)\=-f_x\/f_y$ is obtained.

Hence, the tangent vector is $\mathbf{R}\prime \left(x\right)\=\left[\begin{array}{c}1\\ -f_x\/f_y\end{array}\right]$, and $\nabla f·\mathbf{R}\prime \left(x\right)\=\left[\begin{array}{c}{f}_x\\ f_y\end{array}\right]·\left[\begin{array}{c}1\\ -f_x\/f_y\end{array}\right]$ = $f_x-f_x\=0$.

The gradient $\nabla f$ is therefore orthogonal to the level curve $y\left(x\right)$ defined implicitly by $f\left(x\,y\right)\=c$.