Помогите решить 29 вариант, не понимаю как решать..

Question 1

Question 2

Решите задачу:

$2)\; \; f(x,y)=cos(x^2-2y)\\\\f'_{x}=-sin(x^2-2y)\cdot 2x=-2x\, sin(x^2-2y)\\\\f''_{xx}=-2\, sin(x^2-2y)-2x\cdot cos(x^2-2y)\cdot 2x=\\\\=-2\, sin(x^2-2y)-4x^2\, cos(x^2-2y)\\\\f'_{xy}=-2x\cdot cos(x^2-2y)\cdot (-2)=4x\cdot cos(x^2-2y)\\\\f'_{y}=-sin(x^2-2y)\cdot (-2)=2\, sin(x^2-2y)\\\\f''_{yy}=2\, cos(x^2-2y)\cdot (-2)=-4\, cos(x^2-2y)\\\\f''_{yx}=f''_{xy}$

$1)\; \; u=cos(xy-z^2)- \frac{\sqrt{x}}{\sqrt{y-z^2}} \\\\u'_{x}=-sin(xy-z^2)\cdot y- \frac{1}{\sqrt{y-z^2}}\cdot \frac{1}{2\sqrt{x}}\\\\u'_{xx}=-y\cdot cos(xy-z^2)\cdot y- \frac{1}{2\sqrt{y-z^2}} \cdot \frac{-1}{2\sqrt{x^3}} =\\\\=-y^2\, cos(xy-z^2)+\frac{1}{4x\sqrt{x(y-z^2)}}\\\\u''_{xy}=-sin(xy-z^2)-y\cdot cos(xy-z^2)\cdot x- \frac{1}{2\sqrt{x}} \cdot \frac{-1}{2\sqrt{(y-z^2)^3}} =\\\\=-sin(xy-z^2)-xy\cdot cos(xy-z^2)+\frac{1}{4\sqrt{x(y-z^2)^3}}$

$u''_{xz}=-y\cdot cos(xy-z^2)\cdot (-2z)-\frac{1}{2\sqrt{x}}\cdot \frac{-1\cdot (-2z)}{2\sqrt{(y-z^2)^3}} =\\\\=-y\cdot cos(xy-z^2)- \frac{z}{2\sqrt{x(y-z^2)^3}}$

$u'_{y}=-sin(xy-z^2)\cdot x-\sqrt{x}\cdot \frac{-1}{\sqrt{(y-z^2)^3}} =\\\\=-x\cdot sin(xy-z^2)+\sqrt{x}\cdot \frac{1}{2\sqrt{(y-z^2)^3}} \\\\u'_{yy}=-x\cdot cos(xy-z^2)\cdot x+\sqrt{x}\cdot \frac{1}{2}\cdot \frac{-3}{2\sqrt{(y-z^2)^5}} =\\\\=-x^2\cdot cos(xy-z^2)- \frac{3\sqrt{x}}{4\sqrt{(y-z^2)^5}} \\\\u'_{yz}=-x\cdot cos(xy-z^2)\cdot (-2z)+\frac{\sqrt{x}}{2}\cdot \frac{-3\cdot (-2z)}{2\sqrt{(y-z^2)^5}} =\\\\=2xz\cdot cos(xy-z^2)+ \frac{3z\sqrt{x}}{2\sqrt{(y-z^2)^5}} \\\\u'_{yx}=u'_{xy}$

$u'_{z}=-sin(xy-z^2)\cdot (-2z)-\sqrt{x}\cdot \frac{-1\cdot (-2z)}{2\sqrt{(y-z^2)^3}} =\\\\=2z\cdot sin(xy-z^2)- \sqrt{x}\cdot \frac{z}{\sqrt{(y-z^2)^3}}$

$u''_{zz}=2\, sin(xy-z^2)+2z\cdot cos(xy-z^2)\cdot (-2z)-\\\\-\sqrt{x}\cdot \frac{\sqrt{(y-z^2)^3}-z\cdot \frac{3}{2}\cdot (y-z^2)^{1/2}\cdot (-2z)}{(y-z^2)^3} =2sin(xy-z^2)-\\\\-4z^2\cdot cos(xy-z^2)-\sqrt{x}\cdot \frac{\sqrt{(y-z^2)^3}+6z^2\sqrt{y-z^2}}{(y-z^2)^3} =2\, sin(xy-z^2)-\\\\-4z^2\cdot cos(xy-z^2)-\sqrt{x}\cdot \frac{2(y-z^2)+6z^2}{\sqrt{(y-z^2)^5}}$

$u''_{zx}=u''_{xz}\; ,\; \; \; u''_{zy}=u''_{yz}$

Question 3

спасибо