# Finding level curves given that $-y∂f/∂x+x∂f/∂y =0$

The question is:

suppose $f:R^2 \to R$ is a differentiable function whose gradient is nowhere $0$ and that satisfies

$-y\cdot∂f/∂x + x\cdot ∂f/∂y = 0$

1) find the level curves of $f$

2) show that there is a differentiable function $F$ defined on the set of positive real numbers so that $f(X) =F(|x|)$

I'm really quite clueless on how to do this problem. Thanks

For part 1, notice that this equation can also be written as $$[-y,x] \cdot [f_x,f_y]=0$$ We know that the level curves are perpendicular to the gradient, and we now know that the vector $[-y,x]$ is also perpendicular to the gradient. So, the slope of the level curve at a point $(x,y)$ is $-\frac{x}{y}$
We know that the gradient is never 0, so there will never be an ambiguous case, now we just solve the differential equation $\frac{dx}{dy} = -\frac{x}{y}$, which turns out to give us circles. You can probably solve that on your own, so I'll leave you to it.
I don't really understand what is being asked in 2, as you defined f as a function of two variables, but then you write $f(x)$. If you can clarify I might be able to help you further.