The differential equation
$$f(t,x)+g(t,x)\dot x=0\tag{*}$$
with $(t,x)\in U\subset\mathbb R^2$ and $U$ open is called exact, if there is a continuous and differentiable function $F\colon U\to\mathbb R$, s.t. $\frac{\partial F}{\partial t}(t,x)=f(t,x)$ and $\frac{\partial F}{\partial x}(t,x)=g(t,x)$ for $(t,x)\in U$.
Show the following: If $(*)$ is exact then the solutions of $(*)$ are the differentiable functions $x=x(t)$ with graphs in $U$ and for which $F(t,x(t))$ is constant.
Proof.
Let $F\colon U\to\mathbb R$ be continuous and differentiable with $\frac{\partial F}{\partial t}(t,x)=f(t,x)$ and $\frac{\partial F}{\partial x}(t,x)=g(t,x)$. If $x=x(t)$ is a differentiable function with graph in $U$ s.t.
$$F(t,x(t))=c,~c\in\mathbb R\tag{**}$$
then differentiation of both sides of $(**)$ yields
$$\frac{\partial F}{\partial t}(t,x)+\frac{\partial F}{\partial x}(t,x)\dot x=0,$$
hence
$$f(t,x)+g(t,x)\dot x=0$$
and thus $x$ is a solution.
If $x=x(t)$ is a solution of $(*)$ then
$$\frac{\partial F}{\partial t}(t,x)+\frac{\partial F}{\partial x}(t,x)\dot x=0\implies\frac{\mathrm d}{\mathrm dt}\left(F(t,x(t))\right) = 0 \implies F(t,x(t))=c,~c\in\mathbb R.$$
I am really struggling to understand both the problem as well as the proof due to the notational overload, multiple functions and variables which make this really confusing.
In the first part there is apparently a function $F$ that can be derived (using the chain rule) to obtain $(*)$ but why is $x$ called a solution if we are interested in the form of $F$ which should be constant?
Why are the graphs of the functions relevant?
Why is the second part ($x=x(t)\implies F(t,x(t))=c$) even necessary in this context - especially seing as $x$ is seemingly something arbitrary.
Because we need to show two separate things (see my answer to your third question).
The graphs are not relevant, but it is necessary that you can compute $F$ and since its domain is $U$, the graphs must be in $U$.
Because you need to show to things:
that each solution of $(*)$ is a differentiable function $x=x(t)$ with graph in $U$ and for which $F(t,x(t))$ is constant;
that each differentiable functions $x=x(t)$ with graph in $U$ and for which $F(t,x(t))$ is constant is a solution of $(*)$.