An example of an exact differential equation where the function F is
F= 1/2 ((x^2)+(y^2) + 10  ((x^2)*(y^2) ))

The equation is

x(1+ 10 (y^2)) dx + dy (1+ 10 (x^2))=0

Since P = x(1+10 (y^2)) and Q=y (1+ 10 (x^2))
we see that

 (Subscript[P, y])=20 xy

also 

(Subscript[Q, x])=20 xy   =(Subscript[P, y])

Hence this is an exact differential equation. We can find the solution by setting 
P= (Subscript[F, x])
and Q= (Subscript[F, y])   

where F can be found from integrating these equations
F= \integral dx P  = 1/2 (x^2)(1 + 10 (y^2)) + A(y)   (Here A is an arbitrary function of y. It is allowed since the integration over x assumes that y is a constant)

Similarly integrating Q
F= \integral dy Q= 1/2 (y^2)(1 + 10 (x^2)) + B(x)

Equating the two we get

1/2 (y^2)(1 + 10 (x^2)) + B(x)= 1/2 (x^2)(1 + 10 (y^2)) + A(y)

This equation is easily soved by choosing A(y)= 1/2 (y^2)    and B(x)=1/2 x^2

Plugging in, we get

F=1/2 ((x^2)+ (y^2)+10 (x^2)(y^2))

Summarizing, our differential equation 
x(1+ 10 (y^2)) dx + dy (1+ 10 (x^2))=0 
is completely equivalent to saying that
F(x,y)= (Subscript[C, 0]), 
where   (Subscript[C, 0]) is an arbitrary  constant and F is defined above.

To understand the meaning of this constant, and also of F, we have spoken of F as a height of a mountain, and the value of (Subscript[C, 0]) as specifying the height at which we are seeking a (x,y) curve. Let us see this in pictures.

f=1/2(x^2+y^2 +10x^2 y^2)

Plot3D[f,{x,-5,5},{y,-5,5}]

We observe that the mountain has a steep climb away from (0,0), which the bottom of the valley.

We can now specify the height  (Subscript[C, 0]) as we like and discover the curves at that height.

ContourPlot[f==10,{x,-5,5},{y,-5,5}, FrameLabel->"Height (Subscript[C, 0])= 10"]

ContourPlot[f==20,{x,-5,5},{y,-5,5}, FrameLabel->"Height (Subscript[C, 0])= 20"]

ContourPlot[f==500,{x,-5,5},{y,-5,5}, FrameLabel->"Height (Subscript[C, 0])= 500"]