A surface is represented by the following multivariable function, f(x,y)=x^3+y^3-3x-3y+1 Calculate coordinates of stationary points.

Question
Multivariable functions
A surface is represented by the following multivariable function,
$$\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{3}}+{y}^{{3}}-{3}{x}-{3}{y}+{1}$$
Calculate coordinates of stationary points.

2021-02-19
Obtain $$\displaystyle{f}_{{x}}{\quad\text{and}\quad}{f}_{{y}}$$ as follows.
$$\displaystyle{{f}_{{x}}{\left({x},{y}\right)}}=\frac{\partial}{{\partial{x}}}{\left({x}^{{3}}+{y}^{{3}}-{3}{x}-{3}{y}+{1}\right)}$$
$$\displaystyle={3}{x}^{{2}}+{0}-{3}-{0}+{0}$$
$$\displaystyle={3}{x}^{{2}}-{3}$$
$$\displaystyle{{f}_{{y}}{\left({x},{y}\right)}}\frac{\partial}{{\partial{y}}}{\left({x}^{{3}}+{y}^{{3}}-{3}{x}-{3}{y}+{1}\right)}$$
$$\displaystyle={0}+{3}{y}^{{2}}-{0}-{3}+{0}$$
$$\displaystyle={3}{y}^{{2}}-{3}$$
Now obtain $$\displaystyle{f}_{{x}}={0}{\quad\text{and}\quad}{f}_{{y}}={0}$$ as follows.
$$\displaystyle{f}_{{x}}={0}$$
$$\displaystyle\Rightarrow{3}{x}^{{2}}-{3}={0}$$
$$\displaystyle{3}{\left({x}^{{2}}-{1}\right)}={0}$$
$$\displaystyle{x}^{{2}}-{1}={0}$$
$$\displaystyle{x}^{{2}}={1}$$
$$\displaystyle{x}=\sqrt{{1}}$$
$$\displaystyle{x}=\pm{1}$$
$$\displaystyle{f}_{{y}}={0}$$
$$\displaystyle\Rightarrow{3}{y}^{{2}}-{3}={0}$$
$$\displaystyle{3}{\left({y}^{{2}}-{1}\right)}={0}$$
$$\displaystyle{y}^{{2}}-{1}={0}$$
$$\displaystyle{y}^{{2}}={1}$$
$$\displaystyle{y}=\sqrt{{1}}$$
$$\displaystyle{y}=\pm{1}$$
Thus, the coordinates of stationary points are $$\displaystyle{P}_{{1}}{\left({1},{1}\right)},{P}_{{2}}{\left({1},−{1}\right)},{P}_{{3}}{\left(−{1},{1}\right)},{\quad\text{and}\quad}{P}_{{4}}{\left(−{1},−{1}\right)}$$.

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