Найти дифференциал функции указанного порядка

Решите задачу:

$u=\frac yx\\ du=\frac{\partial^3u}{\partial x^3}+3\frac{\partial^3u}{\partial x^2\partial y}+3\frac{\partial^3u}{\partial x\partial y^2}+\frac{\partial^3u}{\partial y^3}\\ \frac{\partial u}{\partial x}=-\frac y{x^2}\\ \frac{\partial u}{\partial y}=\frac1x\\ \frac{\partial^2u}{\partial x^2}=\frac{2y}{x^3}\\ \frac{\partial^2u}{\partial y^2}=0\\ \frac{\partial^2u}{\partial x\partial y}=-\frac1{x^2}\\ \frac{\partial^3u}{\partial x^3}=-\frac{6y}{x^4}\\ \frac{\partial^3u}{\partial x^2\partial y}=\frac2{x^3}$

$\frac{\partial^3u}{\partial x\partial y^2}=0\\ \frac{\partial^3u}{\partial y^2}=0\\d^3u=-\frac{6y}{x^4}+3\cdot\frac2{x^3}+3\cdot0+0=\frac6{x^3}\left(1-\frac yx\right)$