Помогите решить, очень нужно.

Решите задачу:

$z=\sqrt{2x+y^2-1};\, \frac{\partial z}{\partial x}= \frac{1}{\sqrt{2x+y^2-1}};\, \frac{\partial z}{\partial y}= \frac{y}{\sqrt{2x+y^2-1}};\,$
$\frac{\partial^2 z}{\partial x^2}= \frac{\partial}{\partial x} (\frac{1}{\sqrt{2x+y^2-1}})=-\frac{1}{\sqrt{(2x+y^2-1)^3}};\, \frac{\partial^2 z}{\partial x^2}(1;0)=-1;$
$\frac{\partial^2 z}{\partial y^2}=\frac{\sqrt{2x+y^2-1}-\frac{y^2}{\sqrt{2x+y^2-1}}}{2x+y^2-1}=\frac{2x-1}{\sqrt{(2x+y^2-1)^3}};$
$\frac{\partial^2 z}{\partial y^2}(1;0)=1;\,\frac{\partial^2 z}{\partial y \partial x}= \frac{\partial}{\partial y} (\frac{1}{\sqrt{2x+y^2-1}})=-\frac{y}{\sqrt{(2x+y^2-1)^3}};$
$\frac{\partial^2 z}{\partial x \partial y}= \frac{\partial}{\partial x} (\frac{y}{\sqrt{2x+y^2-1}})=-\frac{y}{\sqrt{(2x+y^2-1)^3}};$
$P(x, y)=2\cdot(-1)-3=-5.$