$g(x,y)=f(\sqrt {x^2+y^2})$ for (x,y) is not zero vector. Show that $y\frac{\partial g}{\partial x}=x\frac{\partial g}{\partial y}$.

Question

Suppose $f: \mathbb{R} \to \mathbb{R}$ is differentiable. I think it need Euler's theorem to solve it. But I do not know the homogeneous degree of g. My attempt is $$g(tx,ty)=f(\sqrt {(tx)^2+(ty)^2})=f(|x|\sqrt {x^2+y^2})$$ Another question is can I say g (two variable function) is differentiable since f (one variable function) is differentiable?