The following demonstrates what's a state function and what's not a state function.

${\displaystyle q_{rev}\;}$ is not exact differential for a gas obeying van der Waals' equation, but ${\displaystyle {\frac {q_{rev}}{T}}}$ is as demonstrated below:

${\displaystyle dq_{rev}\;=\left({\frac {\partial U}{\partial T}}\right)_{v}dT+\left[P_{ext}+\left({\frac {\partial U}{\partial V}}\right)_{T}\right]dV}$

We assume quasistatic situation, so ${\displaystyle P_{ext}=P\;}$.

${\displaystyle dq_{rev}\;=\left({\frac {\partial U}{\partial T}}\right)_{v}dT+\left[P+\left({\frac {\partial U}{\partial V}}\right)_{T}\right]dV}$

${\displaystyle =C_{v}dT+\left[P+\left({\frac {\partial U}{\partial V}}\right)_{T}\right]dV}$

${\displaystyle =C_{v}dT+\left[P+\left({\frac {a}{{\overline {V}}^{2}}}\right)\right]dV}$

${\displaystyle =C_{v}dT+\left({\frac {RT}{{\overline {V}}-b}}\right)dV}$

Now, you take the cross partial derivatives.

${\displaystyle \left({\frac {\partial C_{v}}{\partial V}}\right)_{T}=0}$

${\displaystyle \left({\frac {\partial \left({\frac {RT}{{\overline {V}}-b}}\right)}{\partial T}}\right)_{V}={\frac {R}{{\overline {V}}-b}}}$

They are not equal; hence, ${\displaystyle q_{rev}\;}$ is not exact differential (not a state function).

However, if we take ${\displaystyle {\frac {q_{rev}}{T}}}$ it will be exact differential (a state function).

${\displaystyle {\frac {dq_{rev}}{T}}={\frac {C_{v}}{T}}dT+\left({\frac {R}{{\overline {V}}-b}}\right)dV}$

Take the cross partial derivatives.

${\displaystyle \left({\frac {\partial \left({\frac {C_{v}}{T}}\right)}{\partial V}}\right)_{T}=0}$

${\displaystyle \left({\frac {\partial \left({\frac {R}{{\overline {V}}-b}}\right)}{\partial T}}\right)_{V}=0}$

Both are equal making ${\displaystyle {\frac {q_{rev}}{T}}}$ exact differential (a state function).