You can solve the first equation algebraically to get $\displaystyle F'=\frac{r}{p}$. Then differentiate the second equation with respect to r:
$\displaystyle \frac{\partial{p}}{\partial{r}}F+\frac{\partial{F} }{\partial{r}}p-L-r\frac{\partial{L}}{\partial{r}}-\frac{\partial{B}}{\partial{r}}=0$
$\displaystyle \frac{\partial{B}}{\partial{r}}=0$ since they are both independent variables. Using the chain rule, since F is a function of L,
$\displaystyle \frac{\partial{F}}{\partial{r}} = F'\frac{\partial{L}}{\partial{r}} = \frac{r}{p}\frac{\partial{L}}{\partial{r}}$. So
$\displaystyle \frac{\partial{p}}{\partial{r}}F+\frac{r}{p}\frac{ \partial{L}}{\partial{r}}p-L-r\frac{\partial{L}}{\partial{r}}=0$
The second and fourth terms cancel, and so you get $\displaystyle \frac{\partial{p}}{\partial{r}}=\frac{L}{F}$.
I haven't done the other three derivatives, but I would guess that they're similar. If you have the answers already, that usually helps....
- Hollywood