Привет, мне нужна помощь 1 и 3 задание , желательно подробное решение

$1.\;z(x,y)=\frac{\cos(x^2-y^2)}{3y}\\ \frac{\partial z}{\partial x}=-\frac{2x\sin(x^2-y^2)}{3y}\\ \frac{\partial z}{\partial y}=\frac{-(-2y)\sin(x^2-y^2)3y-3\cos(x^2-y^2)}{9y^2}=\frac{2y^2\sin(x^2-y^2)3y-\cos(x^2-y^2)}{3y^2}\\ dz=\left(-\frac{2x\sin(x^2-y^2)}{3y}\right)dx+\left(\frac{2y^2\sin(x^2-y^2)3y-\cos(x^2-y^2)}{3y^2}\right)dy$

$2.\;z(x,y)=x^3-5x^2y^3+y^3+3y\\ \frac{\partial z}{\partial x}=3x^2-10xy^3\\ \frac{\partial z}{\partial y}=-15x^2y^2+3y+3\\ \frac{\partial^2 z}{\partial x^2}=6x-10y^3\\ \frac{\partial^2 z}{\partial y^2}=-30x^2y+3\\ \frac{\partial^2 z}{\partial x\partial y}=-30xy^2\\ 3.\;z(x,y)=x^2-4xy+y^2+12y-2\\ \begin{cases} \frac{\partial z}{\partial x}=0\\ \frac{\partial z}{\partial y}=0 \end{cases}\Rightarrow \begin{cases} 2x-4=0\\ -4x+2y+12=0 \end{cases}\Rightarrow \begin{cases} x=2\\ y=-2 \end{cases}$

$A=\frac{\partial^2 z}{\partial x^2}=2\\ B=\frac{\partial^2 z}{\partial x\partial y}=-4\\ C=\frac{\partial^2 z}{\partial y^2}=2\\ \Delta=A\cdotC-B^2=2\cdot2-(-4)^2=4-16=-8<0$

Экстремума нет.