Найти производную функции. 60 БАЛЛОВ!!!

Решите задачу:

$y=arctg \frac{y}{x} \; \; ,\; \; M(1,3)\; \; ,\; \; \vec{e}=(3,4)\\\\\frac{\partial z}{\partial x}= \frac{1}{1+ \frac{y^2}{x^2} } \cdot (- \frac{y}{x^2} )=- \frac{x^2}{x^2+y^2}\cdot \frac{y}{x^2} =- \frac{y}{x^2+y^2} \\\\ \frac{\partial z}{\partial y} = \frac{1}{1+\frac{y^2}{x^2}} \cdot \frac{1}{x}= \frac{x^2}{x^2+y^2}\cdot \frac{1}{x}= \frac{x}{x^2+y^2} \\\\\frac{\partial z(M)}{\partial x} =-\frac{3}{1+9} =-\frac{3}{10}=-0,3\\\\\frac{\partial z(M)}{\partial y} = \frac{1}{1+9}=\frac{1}{10}=0,1$

$\vec{e}=(3,4)\; \; \Rightarrow \; \; cos \alpha = \frac{x_{e}}{|\vec{e}|} = \frac{3}{\sqrt{3^2+4^2}} = \frac{3}{5} \\\\cos \beta = \frac{y_{e}}{|\vec{e}|} = \frac{4}{\sqrt{3^2+4^2}}= \frac{4}{5}\\\\ \frac{\partial z(M)}{\partial \vec{e}}= \frac{\partial z(M)}{\partial x} \cdot cos \alpha +\frac{\partial z(M)}{\partial y} \cdot cos \beta =- 0,3\cdot \frac{3}{5}+0,1 \cdot \frac{4}{5} =-0,1$