Exactness of a differential form by means of a function Good evening. I'm having some trouble with this exercice:
Determine f$\in C^1(R) $such that $f(0)=1$ and the differential form
$$ \omega =  \frac{2xy(f(x))^2}{1+(f(x))^2} dx - \arctan (f(x)) dy$$
be exact.
I reasoned like this:
For the differential form to be exact it must be closed. So it must hold true
$$f_y \left(\frac{2xy(f(x))^2}{ 1+(f(x))^2}\right) = f_x (- \arctan (f(x))$$
doing the calculations I found
$$\frac{2x (f(x))^2}{(1+(f(x))^2)^2} =-\frac {f'(x)}{1+(f(x))^2}$$
that is
$$f'(x) = - \frac{2xf(x)}{1+f(x)^2}$$
At this point I don't know how to continue to find $ f (x)$.
 A: The function $f\colon \mathbb{R}\rightarrow \mathbb{R};x\mapsto \frac{1}{x^2+1}$ is the unique continuously differentiable function on $\mathbb{R}$ such that $f(0)=1$ holds and such that the $1$-form $\omega$ is exact.
As you note, any exact form is closed. Thus, if $\omega$ is exact, we have
$$\frac{\partial}{\partial y}\Big(\frac{2xy(f(x))^2}{1+f(x)^2}\Big)=  -\frac{\partial}{\partial x} \Big(\operatorname{arctan}(f(x))\Big).$$
Differentiating yields the following ordinary differential equation
$$-2x(f(x))^2=f'(x).$$ Observe that you made a mistake in the calculation of the partial derivatives.
Solving this ODE via separation of variables, as Ted Shifrin suggests, gives for any fixed $C\in \mathbb{R}$ the solution $f(x)=\frac{1}{x^2+C}$. The initial condition $f(0)=1$ then implies $f(x)=\frac{1}{x^2+1}$. Note that the solution to our initial value problem is unique by the Picard-Lindelöf theorem.
In total, given $f(0)=1$, we have seen that the $1$-form $\omega$ is exact only if $f(x)=\frac{1}{x^2+1}$. Conversely, if $f(x)=\frac{1}{x^2+1}$ holds, then $\omega$ is exact. Namely, one convinces oneself that $$F(x,y)=-\operatorname{arctan}\Big(\frac{1}{x^2+1}\Big)y$$ is a primitive of $\omega$ in this case.
