Equation of Partial derivatives

You are correct: the notation $\partial u$, etc, is not really clear. Dividing the equation in the question by $\partial u$ would give, formally, $$ \frac{\partial u}{\partial u}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial u} $$ which doesn't make much sense.

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Here is how I would handle this:

The Chain Rule says $$ \left.\begin{align} \frac{\mathrm{d}u}{\mathrm{d}t}&=\frac{\partial u}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial u}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}\\[3pt] \frac{\mathrm{d}v}{\mathrm{d}t}&=\frac{\partial v}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial v}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t} \end{align}\right\}\hspace{.5cm} \begin{bmatrix}\dfrac{\mathrm{d}u}{\mathrm{d}t}\\\dfrac{\mathrm{d}v}{\mathrm{d}t}\end{bmatrix}=\begin{bmatrix} \dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\ \dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y} \end{bmatrix} \begin{bmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\\dfrac{\mathrm{d}y}{\mathrm{d}t}\end{bmatrix}\tag1 $$ and $$ \left.\begin{align} \frac{\mathrm{d}x}{\mathrm{d}t}&=\frac{\partial x}{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}+\frac{\partial x}{\partial v}\frac{\mathrm{d}v}{\mathrm{d}t}\\[3pt] \frac{\mathrm{d}y}{\mathrm{d}t}&=\frac{\partial y}{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}+\frac{\partial y}{\partial v}\frac{\mathrm{d}v}{\mathrm{d}t} \end{align}\right\}\hspace{.5cm} \begin{bmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\\dfrac{\mathrm{d}y}{\mathrm{d}t}\end{bmatrix}=\begin{bmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\ \dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v} \end{bmatrix} \begin{bmatrix}\dfrac{\mathrm{d}u}{\mathrm{d}t}\\\dfrac{\mathrm{d}v}{\mathrm{d}t}\end{bmatrix}\tag2 $$ Plugging $(1)$ into $(2)$ gives $$ \begin{bmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\\dfrac{\mathrm{d}y}{\mathrm{d}t}\end{bmatrix} =\begin{bmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\ \dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v} \end{bmatrix} \begin{bmatrix} \dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\ \dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y} \end{bmatrix} \begin{bmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\\dfrac{\mathrm{d}y}{\mathrm{d}t}\end{bmatrix}\tag3 $$ That is, $$ \begin{bmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\ \dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v} \end{bmatrix} \begin{bmatrix} \dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\ \dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y} \end{bmatrix} =I\tag4 $$

There's nothing too deep going on here. The point is that if $f : \mathbb{R}^n \to \mathbb{R}^n$ and $g : \mathbb{R}^n \to \mathbb{R}^n$ satisfy $f \circ g = I$, then $Df|_{g(x)}Dg|_{x} = DI|_{x} = I$. All we used was the chain rule and the fact that the Jacobian of the identity transformation is the identity matrix.