Let $\Omega$ be a domain in $\mathbb R^2$ and $C$ be a smooth curve wholly contained in $\Omega$. Moreover, $C$ is parametrized by $x(t), y(t)$. If $Q: \Omega \rightarrow \mathbb R$ has continuous first-order partials, the directional derivative $D_n$ (in the direction of the unit normal) can be written as:
$$D_nu = u_{x} y' - u_{y} x'$$
I don't believe that it makes sense to have $\frac{\partial}{\partial x} D_nu$ or $\frac{\partial}{\partial y} D_nu$. But, if you look at $D_nu$ as a function of $t$, our parameter, then $\frac{\partial}{\partial x} D_nu = \frac{\partial}{\partial y} D_nu = 0$.
This question came up when I was doing the proof to show that a harmonic $u$ is infinitely differentiable. Suppose that $D$ is a Jordan domain inside $\Omega$. Via Green's identity:
$$u(x, y) = \frac{-1}{2\pi} \int_{\partial D} \text{ln}|(x, y) - \sigma(t)| D_nu - u(\sigma(t)) D_n \text{ln}|(x, y) -\sigma(t)| \space dt$$
Since the integrand is continuous, when we differentiate $u$ with respect to $x$, we can take the derivative operation inside the integral to yield .... Well, what do you have in $\frac{\partial}{\partial x} D_nu$?
