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?
Hint:

$x(x^2+y^2)^{-1/2}+yy'(x^2+2y^2)=0$

$y\dfrac{dy}{dx}(x^2+2y^2)=-\dfrac{x}{\sqrt{x^2+y^2}}$

$\dfrac{dx}{dy}=-\dfrac{y(x^2+2y^2)\sqrt{x^2+y^2}}{x}$

Let $t=y^2$ ,

Then $\dfrac{dx}{dy}=\dfrac{dx}{dt}\dfrac{dt}{dy}=2y\dfrac{dx}{dt}$

$\therefore2y\dfrac{dx}{dt}=-\dfrac{y(x^2+2y^2)\sqrt{x^2+y^2}}{x}$

$\dfrac{dx}{dt}=-\dfrac{(x^2+2y^2)\sqrt{x^2+y^2}}{2x}$

$\dfrac{dx}{dt}=-\dfrac{(x^2+2t)\sqrt{x^2+t}}{2x}$

Let $u=x^2$ ,

Then $\dfrac{du}{dt}=2x\dfrac{dx}{dt}$

$\therefore\dfrac{1}{2x}\dfrac{du}{dt}=-\dfrac{(x^2+2t)\sqrt{x^2+t}}{2x}$

$\dfrac{du}{dt}=-(x^2+2t)\sqrt{x^2+t}$

$\dfrac{du}{dt}=-(u+2t)\sqrt{u+t}$

Let $v=-\sqrt{u+t}$ ,

Then $u=v^2-t$

$\dfrac{du}{dt}=2v\dfrac{dv}{dt}-1$

$\therefore2v\dfrac{dv}{dt}-1=(v^2+t)v$

$2v\dfrac{dv}{dt}=v^3+tv+1$

$(vt+v^3+1)\dfrac{dt}{dv}=2v$

This belongs to an Abel equation of the second kind.