Найти производную неявной функции.

Question 1

Найти производную неявной функции. $x^{ \frac{2}{3} } *ln xy+ xy=0$

Question 2

Решите задачу:

$x^{\frac{2}{3}}\cdot\ln( xy)+xy=0,\\\\\left(x^{\frac{2}{3}}\cdot\ln( xy)\right)'=-(xy)',\\\\\left(x^\frac{2}{3}\right)'\ln(xy)+x^{\frac{2}{3}}\left(\ln(xy)\right)'=-\left(x'y+xy'\right),\\\\\frac{2}{3}x^{\frac{2}{3}-1}\cdot\ln(xy)+x^{\frac{2}{3}}\cdot\frac{1}{xy}\cdot(xy)'=-\left(1\cdot y+xy'\right),\\\\\frac{2}{3}x^{-\frac{1}{3}}\cdot\ln(xy)+\frac{x^{\frac{2}{3}}}{xy}\cdot\left(y+xy'\right)=-y-xy',\\\\\frac{2\ln(xy)}{3\sqrt[3]{x}}+\frac{x^{\frac{2}{3}}\left(y+xy'\right)}{xy}=-y-xy',$

$\frac{2\ln(xy)}{3\sqrt[3]{x}}+\frac{y+xy'}{\sqrt[3]{x}y}=-y-xy'\ |\bullet\sqrt[3]{x}y,\\\\\frac{2\ln(xy)\cdot\sqrt[3]{x}y}{3\sqrt[3]{x}}+y+xy'=\sqrt[3]{x}y\left(-y-xy'\rigth),$

$\frac{2y\ln(xy)}{3}+y+xy'=-y^2\sqrt[3]{x}-xyy'\sqrt[3]{x},\\\\\frac{2y\ln(xy)}{3}+y+xy'=-y^2\sqrt[3]{x}-yy'\sqrt[3]{x^4},\\\\xy'+yy'\sqrt[3]{x^4}=-y^2\sqrt[3]{x}-y-\frac{2y\ln(xy)}{3},\\\\y'\left(x+y\sqrt[3]{x^4}\right)=-y\left(y\sqrt[3]{x}+1+\frac{2\ln(xy)}{3}\right),\\\\y'=\frac{-y\left(y\sqrt[3]x+1+\frac{2\ln(xy)}{3}\right)}{x+y\sqrt[3]{x^4}},\\\\y'=\frac{-y\left(y\sqrt[3]x+1+\frac{2\ln(xy)}{3}\right)}{x\left(1+y\sqrt[3]{x}\right)}.$

$OTBET:\ \ y'=\frac{-y\left(y\sqrt[3]x+1+\frac{2\ln(xy)}{3}\right)}{x\left(1+y\sqrt[3]{x}\right)}.$