Question Number 143006 by mathdanisur last updated on 08/Jun/21
$$\mathrm{1}.\:{y}\frac{\partial{z}}{\partial{x}}\:+\:{z}\frac{\partial{z}}{\partial{y}}\:=\:\frac{{y}}{{x}} \\ $$$$\mathrm{2}.\:{x}^{\mathrm{2}} \frac{\partial{z}}{\partial{x}}\:−\:{xy}\frac{\partial{z}}{\partial{y}}\:+\:{y}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{3}.\:\begin{cases}{\frac{\partial{z}}{\partial{x}}\:=\:\frac{{z}}{{x}}}\\{\frac{\partial{z}}{\partial{y}}\:=\:\frac{\mathrm{2}{z}}{{y}}}\end{cases} \\ $$