Let $$H = \mathbb f(S,V) $$ $$\dfrac{\partial H}{\partial S}S\sqrt{V} = \mathbb g(H) $$ $$\dfrac{\partial H}{\partial V}\sqrt{V} = \mathbb h(H) $$
Note that functions $\mathbb g$ and $\mathbb h$ should be expressed purely in terms of $H$, and should not contain either $S$ or $V$. Find an expression for $H$ in terms of $S$ and $V$.
Thanks.
It is supposed that $g(H)$ is a known function. Hense you can express $$G(H)=\int \frac{dH}{g(H)}$$ and the inverse function of $G(H)=x$ can be expressed : $H=G^{-1}(x)$ Then : $$\frac{\partial G}{\partial S}=\frac{1}{g(H)}\frac{\partial H}{\partial S}$$ With $\frac{\partial H}{\partial S}S\sqrt{V}=g(H)$ : $$\frac{\partial G}{\partial S}=\frac{1}{S\sqrt{V}}$$ $$G(H)=\frac{\ln(S]}{\sqrt{V}}+\Phi_1(V)$$ $\Phi_1$ is any derivable function. The result is : $$H=G^{-1}\left(\frac{\ln(S]}{\sqrt{V}}+\Phi_1(V)\right)$$
THE SAME METHOD with $\frac{\partial H}{\partial V}\sqrt{V}=h(H)$ :
It is supposed that $h(H)$ is a known function. Hense you can express $$F(H)=\int \frac{dH}{h(H)}$$ and the inverse function of $F(H)=x$ can be expressed : $H=F^{-1}(x)$ Then : $$\frac{\partial F}{\partial V}=\frac{1}{h(H)}\frac{\partial H}{\partial V}$$
$$\frac{\partial F}{\partial V}=\frac{1}{\sqrt{V}}$$ $$F(H)=2\sqrt{V}+\Phi_2(S)$$ $\Phi_2$ is any derivable function. The result is : $$H=F^{-1}\left(2\sqrt{V}+\Phi_2(S)\right)$$
In fact, we obtain two distinct forms for the function $H(S,V)$. So, there is a condition for consistency : $$G^{-1}\left(\frac{\ln(S]}{\sqrt{V}}+\Phi_1(V)\right)=F^{-1}\left(2\sqrt{V}+\Phi_2(S)\right)$$ This means that a relationship must comply between the functions $g$ and $h$ so that the above relationship complies.