How can i show that this differential equation is exact? The Problem is to show (proof) a given differential equation is exact and to solve the equation.
$$(2y^2+6xy-x^2)\ dx\ +\ (y^2 + 4xy + 3x^2)\ dy = 0$$
My Approach: i can see this is ordinary differential equation, it should be non-linear and of first order. 
I thought i could solve the problem by integrating each part of the equation by self, using the other variable as a constant. this led to the following:
$$(2y^2+6xy-x^2)\ dx\ = -\frac{x^3}{3}+3x^2y+2xy^2$$
$$(y^2 + 4xy + 3x^2)\ dy = 3x^2y+2xy^2+\frac{y^3}{3}$$
So alltogether i would get: 
$$(2y^2+6xy-x^2)\ dx\ +\ (y^2 + 4xy + 3x^2)\ dy = -\frac{x^3}{3}+6x^4y+2xy^2+\frac{y^3}{3}$$
And i have: $$-\frac{x^3}{3}+6x^4y+2xy^2+\frac{y^3}{3}=0$$
But i that cannot be the solution, it seems to be to easy. And additional i don't have a clue to give the Proof, i am looking for. I think i am stuck.
 A: Like you said:
$$
\frac{\partial F(x,y)}{\partial x} = p(x,y), \quad \text{and} \frac{\partial F(x,y)}{\partial y} = q(x,y),
$$
then $F(x,y) = C$ solves 
$$
p(x,y) + q(x,y)\frac{dy}{dx} = 0,
$$
basically we take derivative w.r.t. $x$ in $F(x,y) = C$.
Now your equation reads:
$$
\frac{\partial F(x,y)}{\partial x} = 2y^2+6xy-x^2 \implies F(x,y) = \int ( 2y^2+6xy-x^2) dx + \color{red}{H(y)},
$$
where $H(y)$ is the key, because a function purely based on $y$ is lost when we take $\partial_x$. Then
$$
F(x,y) = 2y^2x +3x^2y - \frac{x^3}{3}+  H(y).\tag{1}
$$
Here we can use the second condition:
$$
 \frac{\partial F(x,y)}{\partial y} = y^2 + 4xy + 3x^2,
$$
plugging (1) into above:
$$
 \frac{\partial }{\partial y}\left(2y^2x +3x^2y - \frac{x^3}{3}+  H(y)\right) = y^2 + 4xy + 3x^2,
$$
for first three terms, take partial derivative w.r.t. $y$. For $H(y)$, which is a function only of $y$-variable, take $\partial/\partial y$ is the same as $d/dy$, hence this gives you (notice some terms get canceled):
$$
\frac{d }{d y} H(y) = \ldots
$$
integrating both w.r.t. $y$ as if you are integrating only a one independent variable ODE, then plugging back to (1) you will get your $F(x,y)$.
