Well, yes. But the list is very big. For your example we have that

$$ \left( \frac{\partial P }{ \partial T } \right)\Big|_{S} = - \left( \frac{\partial P }{ \partial S } \right)\Big|_{T} \left( \frac{\partial S }{ \partial T } \right)\Big|_{P}$$

but the LHS due to Maxwell's relations is the same as $ \left( \frac{\partial S}{ \partial V } \right)\Big|_{P}$. But for this guy we have that

$$ \left( \frac{\partial P }{ \partial V } \right)\Big|_{S} = - \left( \frac{\partial P }{ \partial S } \right)\Big|_{V} \left( \frac{\partial S }{ \partial V } \right)\Big|_{P}$$

and from Maxwell we have that $- \left( \frac{\partial P }{ \partial S } \right)\Big|_{V} = \left( \frac{\partial T }{ \partial V } \right)\Big|_{S} $. Now for the RHS of this we have

$$ \left( \frac{\partial T }{ \partial V } \right)\Big|_{S} = - \left( \frac{\partial T }{ \partial S } \right)\Big|_{V} \left( \frac{\partial S }{ \partial V } \right)\Big|_{T}$$

Finally, again from Maxwell we know that $\left( \frac{\partial S }{ \partial V } \right)\Big|_{T} = \left( \frac{\partial P }{ \partial T } \right)\Big|_{V} = -\frac{\alpha}{\kappa}$ where

$$ \alpha = \frac{1}{V}\left( \frac{\partial V}{\partial T} \right)\Big|_{P} $$

and

$$ \kappa = -\frac{1}{V}\left( \frac{\partial V}{\partial P} \right)\Big|_{T} $$

I hope this helps.