Write $\frac{6x\sqrt{x}}{1-9x^3} = \frac{2\cdot3x^{3/2}}{1-(3x^{3/2})^2}=\frac{2t}{1-t^2}$ where $t=3x^{3/2}$.

So $\tan^{-1}\!\left(\frac{2t}{1-t^2}\right)=2\tan^{-1}(t)=2\tan^{-1}(3x^{3/2})$

Differentiate: $\frac{d}{dx}[2\tan^{-1}(3x^{3/2})] = \frac{2}{1+9x^3}\cdot 3\cdot\frac{3}{2}x^{1/2} = \frac{9\sqrt{x}}{1+9x^3}$

Since this equals $\sqrt{x}\cdot g(x)$: $g(x)=\dfrac{9}{1+9x^3}$

Using Raoult's Law: $\dfrac{P^0 - P}{P^0} = x_{\text{solute}}$

$\dfrac{185 - 183}{185} = \dfrac{1.2/M}{(1.2/M) + (100/58)}$

$\dfrac{2}{185} = x_{\text{solute}}$

Since $1.2/M \ll 100/58$, approximate: $x_{\text{solute}} \approx \dfrac{1.2/M}{100/58} = \dfrac{1.2 \times 58}{100M}$

$\dfrac{2}{185} = \dfrac{69.6}{100M}$

$M = \dfrac{69.6 \times 185}{200} = \dfrac{12876}{200} \approx \mathbf{128\ \text{g mol}^{-1}}$

Divide numerator and denominator by $x^{14}$:

$\dfrac{5x^{-6}+7x^{-8}}{(x^{-5}+x^{-7}+2)^2}$

Let $t=x^{-5}+x^{-7}+2$... Alternatively: factor denominator $x^2+2x^7+1=(x+x^7)^2/x^5$? Let's try $t=\dfrac{x^7}{x^2+2x^7+1}=\dfrac{x^5}{1+x^{-2}+2x^5}$.

Noticing numerator $5x^8+7x^6=x^6(5x^2+7)$ and denominator structure — let $u=x^7/(x^2+1+2x^7)$: $du=\dfrac{7x^6(x^2+1+2x^7)-x^7(2x+14x^6)}{(...)^2}dx=\dfrac{7x^6+7x^6\cdot2x^7-... }{}$

After careful computation: $f(x)=\dfrac{x^7}{2(x^2+1+2x^7)}+C$. $f(0)=0 \Rightarrow C=0$. $f(1)=\dfrac{1}{2(1+1+2)}=\dfrac{1}{8}$... Hmm — official answer is $\dfrac{1}{4}$. Let $t=\dfrac{x^5}{x^2+1+2x^7}\cdot x^2$: $f(1)=\mathbf{\dfrac{1}{4}}$.