= Solution
The expansion about the <liquid-gas critical point> is
$$
\bar p=1+4t-6t\phi+9t\phi^2-\frac32\phi^3-\frac{27}{2}t\phi^3+O(\phi^4)+O(t\phi^4).
$$
Differentiating at fixed $t$ gives the requested $-9\phi^2/2+O(\phi^3)+t[-6+O(\phi)]$. At coexistence $\phi=O(\sqrt{-t})$, so to leading nontrivial order keep the cubic odd part $\bar p=1+4t-6t\phi-3\phi^3/2$.
Let $a=\phi_l$, $b=\phi_g$ for this leading cubic model. Equality of endpoint <pressures> gives $a^2+ab+b^2=-4t$. The Maxwell condition in <pressure> coordinates is $\int_a^b(1+\phi)\bar p'(\phi)\,d\phi=0$. Since the endpoint <pressures> agree, its first term vanishes, leaving
$$
(b^2-a^2)\left[-3t-\frac98(a^2+b^2)\right]=0.
$$
If $a+b\ne0$, these two equations would imply $(a-b)^2=0$, contrary to distinct phases. Therefore $a=-b$, and the <pressure> equality gives $a^2=b^2=-4t$. Hence
$$
\boxed{\phi_l=-2\sqrt{-t}+O(t),\qquad\phi_g=2\sqrt{-t}+O(t),\qquad
\bar v_g-\bar v_l\sim4\sqrt{\frac{T_c-T}{T_c}}.}
$$
Thus the coexistence gap has exponent one half. \b[The printed equality $\phi_l(t)=-\phi_g(t)$ is a leading-order statement, not an exact identity for the full <Van der Waals equation>.] Discarded asymmetric terms shift both phase <volumes> at higher order. This distinction is needed even though the leading cubic calculation has exact antisymmetry within its truncation.
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