Let $A=\{(x,y)\in\mathbb{R^2}:x>0\}$. Let $f:A\longrightarrow{\mathbb{R}}$ differentiable satisfying: $$x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=0$$
How can I prove the function $h:\mathbb{R}\rightarrow\mathbb{R}$ differentiable so that $f(x,y)=h(\frac{y}{x})$ exists?
Define the function $g:(x,z) \mapsto f(x,xz).$ Since $f$ is differentiable, so is $g$.
To make notation clear let $D_1f $ and $D_2f$ denote partial derivatives with respect to the first and second argument for $f$. The hypothesis is
$$\tag{*}x_1 D_1f(x_1,x_2) + x_2 D_2f(x_1,x_2) = 0.$$
We have using the chain rule
$$\frac{\partial g}{\partial x}(x,z) = D_1f(x,xz) + zD_2f(x,xz).$$
Since $x > 0$ in $A$, we can write this as follows and apply (*) to obtain
$$\frac{\partial g}{\partial x}(x,z) = \frac{1}{x}(xD_1f(x,xz) + zxD_2f(x,xz)) = 0.$$
Thus, $g$ is independent of $x$. Defining $h(z) = g(x,z) = f(x,xz)$ we have
$$h(y/x) = f(x,x(y/x)) = f(x,y).$$