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

A surface is represented by the following multivariable function,
$f\left(x,y\right)=\frac{1}{3}{x}^{3}+{y}^{2}-2xy-6x-3y+4$
a) Calculate ${f}_{xx},{f}_{yx},{f}_{xy}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{f}_{yy}$
b) Calculate coordinates of stationary points.
c) Classify all stationary points.

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