Prove that the form $f(x^2+y^2)x\,dx +f(x^2+y^2)y\,dy $ is closed for a continuous $f$. I'm not even sure if the exterior derivative exists in this case, but I need to prove that the integral of the form over any closed path is equal to $0$.
I tried using a pullback to polar coordinates $\phi(r,\theta) = (r\cos\theta, r\sin\theta)$ and got that the integral over a closed path is:
$$\int f(r^2)rd \text{ ??}$$
This is after simplification. I still don't understand formally how the pullback should look, and I think I did it wrong. I don't know what's supposed to go after the $d$, hence the ??.
Any help would be much appreciated!
Edit: As Tsemo Aristide pointed, we prove below that the vector field is conservative. If by closed you mean a closed differential form, it is easy to come with examples of continuous functions $f$ for which $f(x^2+y^2)x$ is not differentiable with respect to $x$, and any such example would be a counterexample to the problem. Indeed
$$\lim_{x \to a} \frac{f(x^2+y_0^2)x - f(a^2+y_0^2)a}{x-a}=\lim_{x \to a} \frac{f(x^2+y_0^2)x - f(x^2+y_0^2)a}{x-a}+\lim_{x \to a} \frac{f(x^2+y_0^2)a - f(a^2+y_0^2)a}{x-a}$$
As the first limit exists by the continuity of $f$, all you need is to make sure $a \neq 0$ and that $f(x^2+\alpha)$ is not differentiable at $a$ for a fixed $\alpha$.
Hint Let $F$ be an antiderivative of $f$, and define the vector field
$$G(x,y) =\frac{1}{2}F(x^2+y^2)$$
What is the gradient of $G$?