Exact ODE: show that $y$ is a solution iff it is in a level set of $F$ 
A Differential equation of the form
$p(x,y(x))\dot{y}(x)+q(x,y(x))=0\hspace{1cm}(1.1)$
is called exact, if there is a differentiable function $\mathbb{R}^2\mapsto\mathbb{R}$ such that
$p(x,y(x))=\dfrac{\partial F(x,y)}{\partial y}\hspace{2cm} q(x,y)=\dfrac{\partial F(x,y)}{\partial x}$
Show that $y$ is a solution of $(1.1)$ iff it is in a level set of $F$.

$"\Rightarrow"$
Let $y$ be a solution of $(1.1)\Rightarrow p(x,y)\dot{y}=-q(x,y)\hspace{1cm}(1.2)$
$\hspace{4,8cm}\Rightarrow\dfrac{\partial F(x,y)}{\partial y}\dot{y}=-\dfrac{\partial F(x,y)}{\partial x}$
What I would now do is treat the partial derivatives as fraction and get that
$\hspace{4,8cm}-\dfrac{\partial y}{\partial x}=\dot{y}$
Which would mean that the derivative $\dot{y}$ is zero so $y$ is constant an therefore in a level set of $F$
I'm well aware that this is quite naive and would therefore like to know if the idea is atleast correct and how I can make it formally correct if so.
$"\Leftarrow"$
Let $y$ be an element of level set $L_F(c)$.
$\Rightarrow\exists c\in\mathbb{R}:F(x,y) = c$
$\hspace{6,8cm}\Rightarrow p(x,y)=q(x,y)=0$
and so $y$ is a solution for $(1.1)$
Here I'm quite sure that it's correct.
 A: If $y(x)$ is a solution to the equation
$p(x, y(x))\dot y(x) + q(x, y(x)) = 0, \tag 1$
with
$p(x, y(x)) = \dfrac{\partial F(x, y)}{\partial y}, \tag 2$
$q(x, y(x)) = \dfrac{\partial F(x, y)}{\partial x}, \tag 3$
then the curve
$\gamma(x) = (x, y(x)) \tag 4$
with tangent vector
$\dot \gamma(x) = (1, \dot y(x)) \tag 5$
satisfies the equation
$\dot \gamma(x) \cdot (q(x, y(x)), p(x, y(x)))$
$= (1, \dot y(x)) \cdot (q(x, y(x)), p(x, y(x))) = q(x, y(x)) + p(x, y(x))\dot y(x) = 0; \tag 6$
in the light of (2) and (3) this may be written
$\dot \gamma(x) \cdot \nabla F(x, y)$
$= (1, \dot y(x)) \cdot \nabla F(x, y) = (1, \dot y(x)) \cdot \left ( \dfrac{\partial F(x, y)}{\partial x}, \dfrac{\partial F(x, y)}{\partial y} \right ) = 0, \tag 7$
and since
$\dfrac{dF(x, y(x))}{dx} = (1, \dot y(x)) \cdot \nabla F(x, y(x)) = \dot \gamma(x) \cdot \nabla F(\gamma(x)), \tag 8$
we see that
$\dfrac{dF(x, y(x))}{dx} = 0, \tag 9$
that is, the curve $\gamma(x) = (x, y(x))$ lies in a set of constant $F(x, y)$, also known as a level set of $F$.
Now suppose that $y(x)$ is a differentiable function of $x$ which satisfies (9); then via (8) we see that (7) binds, and thus that
$(1, \dot y(x)) \cdot \left ( \dfrac{\partial F(x, y)}{\partial x}, \dfrac{\partial F(x, y)}{\partial y} \right ) = 0, \tag{10}$
or
$\dfrac{\partial F(x, y)}{\partial x} + \dfrac{\partial F(x, y)}{\partial y} \dot y(x) = 0, \tag{11}$
and by means of (2)-(3) we immediately arrive at (1), which we see that $y(x)$ satisfies.  Note that we have stipulated that $y(x)$ be differentiable in order to ensure the existence of $\dot y(x)$, essential if the argument is to make sense at all.  In any event, that differentiable $y(x)$ is a solution of (1) if and only if it's graph lises in a level set of $F(x, y)$ has been established.
