Let $f:\mathbb R \to \mathbb R$ a function of class $C^1$, and let
$g(x,y)=f\left(\dfrac{x-y}{x+y}\right)$ for all $x \neq -y$
Prove that the direction of greatest increase of $g$ at $(x_0,y_0)$ with $x_0 \neq -y_0$ is perpendicular to the vector $(x_0,y_0)$.
The attempt at a solution
I know that the direction of greatest increase is $\nabla g(x_0,y_0)$, so, I've tried to show that $<\nabla g(x_0,y_0),(x_0,y_0)>=0$
Now, if I consider $h:\mathbb R^2 \to \mathbb R$ defined as $h(x,y)=\dfrac{x-y}{x+y}$, then $g=f \circ h$.
I have basic problems to calculate $\nabla g$, I know I have to apply the chain rule, but I don't know how to apply it in this particular problem.
Solution after user25004's help:
By the chain rule, $\nabla g(x,y)=f'(h(x,y))\nabla h(x,y)=f'(\dfrac{x-y}{x+y})(\dfrac{2y}{(x+y)^2},\dfrac{-2x}{(x+y)^2})$. Evaluating at $(x_0,y_0)$, I get that
$<\nabla g(x_0,y_0),(x_0,y_0)>=<f'(\dfrac{x_0-y_0}{x_0+y_0})(\dfrac{2y_0}{(x_0+y_0)^2}\dfrac{-2x_0}{(x_0+y_0)^2}),(x_0,y_0)>$
Now, this last expression equals to $f'(\dfrac{x_0-y_0}{x_0+y_0})(\dfrac{2y_0}{(x_0+y_0)^2}x_0 + \dfrac{-2x_0}{(x_0+y_0)^2}y_0)=0$, this proves $<\nabla g(x_0,y_0),(x_0,y_0)>=0$
