Let $h(x,y)=f(\frac{y}{x})+xg(\frac{y}{x})$, then what is the simplest way for calculating the following second order partial derivative: $$x^2\frac{\partial^2 h}{\partial x^2} +y^2\frac{\partial^2 h}{\partial y^2}$$
I think the Euler's Theorem (for Homogeneous functions) would be helpful but I don't know how to apply it.
By the Chain and Product Rules, $$\frac{\partial h}{\partial x}=g\left(\frac yx\right)-\frac y{x^2}\left(f'\left(\frac yx\right)+xg'\left(\frac yx\right)\right)$$ and $$\frac{\partial ^2h}{\partial x^2}=\frac{2y}{x^3}\left(f'\left(\frac yx\right)+xg'\left(\frac yx\right)\right)-\frac y{x^2}\left(-\frac y{x^2}\left(f''\left(\frac yx\right)+xg''\left(\frac yx\right)+2g'\left(\frac yx\right)\right)\right)$$ so after simplifying we have $$x^2\frac{\partial^2h}{\partial x^2}=\frac{2y}xf'\left(\frac yx\right)+\frac{y^2}{x^2}f''\left(\frac yx\right)+\frac{y^2}xg''\left(\frac yx\right)$$ Similarly, $$\frac{\partial h}{\partial y}=\frac1xf'\left(\frac yx\right)+g'\left(\frac yx\right)+g\left(\frac yx\right)$$ and $$\frac{\partial^2h}{\partial y^2}=-\frac1{x^2}f'\left(\frac yx\right)+\frac1{x^2}f''\left(\frac yx\right)+\frac1x\left(g'\left(\frac yx\right)+g''\left(\frac yx\right)\right)$$ so after simplifying we have $$y^2\frac{\partial^2h}{\partial y^2}=-\frac{y^2}{x^2}f'\left(\frac yx\right)+\frac{y^2}{x^2}f''\left(\frac yx\right)+\frac{y^2}xg'\left(\frac yx\right)+\frac{y^2}xg''\left(\frac yx\right)$$ Adding gives $$\boxed{x^2\frac{\partial^2h}{\partial x^2}+y^2\frac{\partial^2h}{\partial y^2}=Af'\left(\frac yx\right)+Bf''\left(\frac yx\right)+Cg'\left(\frac yx\right)+Dg''\left(\frac yx\right)}$$ where $$\boxed{A=\frac{2y}x-\frac{y^2}{x^2},\,B=\frac{2y^2}{x^2},\,C=\frac{y^2}x,\,D=\frac{2y^2}x}$$