How to prove $\oint_{\partial D} \left( u \frac{d u}{d x} dx - u \frac{d u}{d y} dy \right) = \oint_{\partial D}u\frac{d u}{d \mathbf{n}} ds$?

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For some simply connected region $D \subset \mathbb{R}^2$ with boundary $\partial D$, length differential along boundary $ds$ and normal $\mathbf{n}$, and some sufficiently smooth function $u(x,y)$ defined in $\bar{D}$, I need to show

$$\oint_{\partial D} \left( u \frac{\partial u}{\partial x} dx - u \frac{\partial u}{\partial y} dy \right) = \oint_{\partial D} u \frac{\partial u}{\partial \mathbf{n}} ds $$

I know that $\frac{\partial u}{\partial \mathbf{n}} = \mathbf{n}.\nabla u $ and $ds = \sqrt{\left( \frac{dx}{dt} \right) ^2 + \left( \frac{dy}{dt} \right) ^2} dt$ for some parameter $t$. Do I need to use some of Green's identities?

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If $c :[0,l]\rightarrow \partial D$ is one-to-one and onto and $|c'|=1,\ c(s)=(x,y)(s)$, then $$ c'(s)\cdot (-y',x')=0 \Rightarrow {\bf n}=(-y',x')$$

Then $$ \int_c \nabla f \cdot {\bf n} = \int [f_x(-y') + f_y(x') ] ds =\int [-f_x dy + f_y dx ] $$

If we let $f=u^2$ then we obtain an equality in OP