[m]\frac{\partial z}{\partial x}=\frac{\partial (\frac{y}{x})}{\partial x}=(\frac{y}{x})`_{x}=y\cdot (\frac{1}{x})`=-\frac{y}{x^2}[/m]
[m]\frac{\partial z}{\partial y}=\frac{\partial (\frac{y}{x})}{\partial y}=(\frac{y}{x})`_{y}= \frac{1}{x}\cdot(y)`=\frac{1}{x}[/m]
Находим частные производные второго порядка:
[m]\frac{\partial^2 z}{\partial x^2}=\frac{\partial(\frac{\partial z}{\partial x}) }{\partial x}=(z`_{x})`_{x}=-(\frac{y}{x^2})`_{x}=\frac{2y}{x^3}[/m]
[m]\frac{\partial^2 z}{\partial x \partial y}=\frac{\partial(\frac{\partial z}{\partial x}) }{\partial y}=(z`_{x})`_{y}=-(\frac{y}{x^2})`_{x}=-\frac{1}{x^2}[/m]
[m]\frac{\partial^2 z}{\partial y^2}=\frac{\partial(\frac{\partial z}{\partial y}) }{\partial y}=(z`_{y})`_{y}=-(\frac{1}{x})`_{y}=0[/m]
Подставляем в уравнение:
x^2*[m]\frac{2y}{x^3}[/m]+2x*y*([m]-\frac{1}{x^2}[/m])+y^2*0=0 - верно
Значит удовлетворяет