A surface is represented by the following multivariable function,f(x,y)=1/3x^3+y^2-2xy-6x-3y+4a)

Since, we have been given a surface that is represented by the following multivariable function given below:
${\displaystyle f(x,y)=\frac{1}{3}{x}^3+y^2-2xy-6x-3y+4}$ (i)
To calculate:
a) ${\displaystyle f_{xx},f_{yx},f_{xy}\text{and}{f}_{yy}}$
Solution ${\displaystyle \Rightarrow f_x=\frac{3}{3}{x}^2-2y-6}$
${\displaystyle \Rightarrow f_{xx}=2x}$
${\displaystyle f_y=2y-2x-3}$
${\displaystyle f_{yy}=2}$
${\displaystyle f_{xy}=-2}$
${\displaystyle \text{and}{f}_{yx}=-2}$
b) ${\displaystyle f_x=x^2-2y-6}$
${\displaystyle \text{and}{f}_y=2y-2x-3}$
${\displaystyle f_x=0\Rightarrow x^2-2y=6}$ (ii)
${\displaystyle \text{and}{f}_y=0\Rightarrow 2y-2x=3}$ (iii)
Adding equation (ii) and (iii), we have:${\displaystyle -x^2-2x=9}$
Solving for x, we have:
${\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2}a\Rightarrow x=\frac{2\pm \sqrt{4+36}}{2}\Rightarrow x=\frac{2\pm \sqrt{10}}{2}\Rightarrow x=1\pm \sqrt{10}}$
Since, ${\displaystyle 2y-2x=3\Rightarrow 2y=2x+3\Rightarrow y=\frac{2x+3}{2}=\frac{2(1\pm \sqrt{10})+3}{2}=\frac{2\pm 2\sqrt{10}+3}{2}}$
${\displaystyle \Rightarrow y=\frac{1}{2}(5\pm 2\sqrt{10})}$
So, stationary points are:
${\displaystyle (1-\sqrt{10},\frac{5-2\sqrt{10}}{2}),(1+\sqrt{10},\frac{5-2\sqrt{10}}{2}),(1-\sqrt{10},\frac{5+2\sqrt{10}}{2}),(1+\sqrt{10},\frac{5+2\sqrt{10}}{2})}$
c) ${\displaystyle (1+\sqrt{10},\frac{5+2\sqrt{10}}{2})}$ is minimal and ${\displaystyle (1-\sqrt{10},\frac{5-2\sqrt{10}}{2})}$ is a saddle point.