Solving for gas law partial derivative I don't know where to begin with this question.


  A scientist discovered that the state of the unknown gas can be well described by the equation of state below:
  $$P = \frac{RT}{\overline{V}} + \left(\frac{a + bT}{\overline{V}^2}\right)$$
  Find the partial derivative $\left(\frac{\partial \overline{V}}{\partial T}\right)_P$ which is expressed in terms of $\overline{V}, R, b$, and $P$. (Hint: If $\overline{V}$ is a function of two variables $T$ and $P$ (i.e. $\bar{V} = f(T, P)$), then the total differential can be expressed as
  $$d\bar{V} = df(T, P) = \left(\frac{\partial \overline{V}}{\partial T}\right)_PdT + \left(\frac{\partial \overline{V}}{\partial P}\right)_TdP$$
  Use a particular condition to derive the expression $\left(\frac{\partial \overline{V}}{\partial T}\right)_P$ in terms of $\left(\frac{\partial \overline{V}}{\partial P}\right)_T$ and $\left(\frac{\partial P}{\partial T}\right)_{\overline{V}}$.)


This problem is related to the physical chemistry of the gas law. What is the correct approach?
Think of it this way.
$$dP = \left(\partial P\over\partial T\right)_VdT+ \left(\partial P\over\partial V\right)_TdV$$
Now, we want P to be constant (that is, $dP=0$). What does that entail? A certain relation between $dV$ and $dT$ (which, BTW, are now $\partial$ rather than $d$):
$$\left(\partial V\over\partial T\right)_P=-\left(\partial P\over\partial T\right)_V\left/\left(\partial P\over\partial V\right)_T\right.$$