Show that if $u=g(x,y)$ and $v=h(x,y)$ then $\frac{\partial x}{\partial u} = \frac{\frac{\partial v}{\partial y}}{J}$

Question

where J is the Jacobian $\frac{\partial (u,v)}{\partial (x,y)}$

First:

$$ \frac{\frac{\partial v}{\partial y}}{J} = \frac{\frac{\partial v}{\partial y}}{\frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial u}{\partial y} \frac{\partial v}{\partial x}} $$

I do this:

$$ \frac{\frac{\partial v}{\partial y}}{J} = \frac{1}{\frac{\partial u}{\partial x} - \frac{\frac{\partial u}{\partial y} \frac{\partial v}{\partial x}}{\frac{\partial v}{\partial y}}} $$

I then expand this: $$ \frac{\partial u}{\partial y} \frac{\partial v}{\partial x} = \frac{\partial u}{\partial y} \frac{\partial v}{\partial y} \frac{\partial y}{\partial x} $$

Then the above becomes:

$$ \frac{\frac{\partial v}{\partial y}}{J} = \frac{1}{\frac{\partial u}{\partial x} - \frac{\partial u}{\partial y} \frac{\partial y}{\partial x}} $$

But following this logic (which is clearly wrong):

$$ \frac{\partial u}{\partial y} \frac{\partial y}{\partial x} = \frac{\partial u}{\partial x} $$

And so:

$$ \frac{\frac{\partial v}{\partial y}}{J} = \frac{1}{\frac{\partial u}{\partial x} - \frac{\partial u}{\partial x}} = \frac{1}{0} $$