Derivative relation between two equal functions I am stuck with the following problem. Suppose $g: \Bbb R\rightarrow\Bbb R$ is $C^1$. $f(x,y)=g(x^2+y^2)$. I need to show that $xf_y=yf_x$
My attempt was:
$f_x=g_x \cdot 2x$ (1) and $f_y=g_y\cdot 2y$ (2)
Multiplying (1) by y and (2) by x, I get
$yf_x=g_x \cdot2xy$ and $xf_y=g_y\cdot 2xy$
So the only thing now I need to show is that $g_x=g_y$ Does $C^1$ satisfy that condition? If not, how can I prove that? A good hint would be appreciated. I would be very thankful :)
Thanks in advance!

Apply the chain rule
$$f_x(x,y)=g'(x^2+y^2)\cdot (2x) \quad\Longrightarrow\quad y\cdot f_x(x,y)=g'(x^2+y^2)\cdot 2xy$$
$$f_y(x,y)=g'(x^2+y^2)\cdot (2y) \quad\Longrightarrow\quad x\cdot f_y(x,y)=g'(x^2+y^2)\cdot 2xy$$
And the equality follows.
