Question: By eliminating $u$ from the equations, $x=u+v$ and $y=uv^2$, we get that $F(x,y,v)=0$. Where $v$ implicitly defines $x$ as function of $y$ ($v=h(x,y)$). Prove that $\dfrac{\partial h}{\partial x}=\dfrac{h(x,y)}{3h(x,y)-2x}$.

Honestly I have no idea how to start. First, how do we get $F(x,y,v)=0$? What does eliminating means? To give a specific value to $v$? (I wrote it out as the problem was given)

Answer: 
This is certainly not an exact differential equation. Moreover, I don't think it has closed-form solutions at all: Maple doesn't find any. Are you sure you don't have a typo?
Obviously $x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$ isn't an exact differential equation.

So, there is a typo. Moreover the equation is certainly simple. The simplest guess is :
$$x(x^2+y^2)+yy\prime(x^2+2y^2)=0$$
$$x(x^2+y^2)dx+y(x^2+2y^2)dy=0$$
If it is the exact differential of $F(x,y)$ then :

$\begin{cases}
\frac{\partial F}{\partial x}=x(x^2+y^2) \quad\to\quad F=\frac{x^4}{4}+\frac{x^2y^2}{2}+f(y)\\
\frac{\partial F}{\partial y}=y(x^2+2y^2) \quad\to\quad F=\frac{x^2y^2}{2}+2\frac{y^4}{4}+g(x)
\end{cases}$

which implies $f(y)=\frac{y^4}{2}$ and $g(x)=\frac{x^4}{4}$

$F=\frac{x^4}{4}+\frac{x^2y^2}{2}+\frac{y^4}{2}=\text{constant}\quad$ because $dF=0$

$$x^4+2x^2y^2+2y^4=c$$
Solving it for $y$ leads to :
$$y(x)=\pm\sqrt{\frac{-x^2\pm \sqrt{2c-x^4}}{2}}$$