Finding $\frac{dr}{du}$ and $\frac{dr}{dv}$ from $(1-\frac{r}{2m})\,\exp(\frac{r}{2m})=uv$ I have the following relation
$$\left( 1-\dfrac{r}{2m} \right)\, \exp\left( \dfrac{r}{2m} \right)= u v$$
that defines implicitly $r$ as a function of $u$ and $v$. I need to find $\dfrac{dr}{du}$ and $\dfrac{dr}{dv}$.  This is what I have done so far:
I have derived in $u$ and got:
$$\dfrac{1}{2m} \dfrac{dr}{du} \left[ -\exp\left( \dfrac{r}{2m} \right)+\left( 1-\dfrac{r}{2m} \right)\, \exp\left( \dfrac{r}{2m} \right) \right]=v$$
The second term in the brackets is $uv$, so I have:
$$\dfrac{1}{2m}\dfrac{dr}{du}\left[ -\exp\left( \dfrac{r}{2m} \right) +uv \right]=v$$
How do I eliminate the exponent inside the brackets to get a function of only $u$ and $v$?


Comment to the @Avocaddo's post. The chain rule suggests:
$\frac{\partial R}{\partial u} = \frac{\partial R}{\partial r} \cdot \frac{\partial r}{\partial u} + \frac{\partial R}{\partial u} = 0 \Rightarrow \frac{\partial r}{\partial u} = - \frac{\frac{\partial R}{\partial u}}{\frac{\partial R}{\partial r}}$
