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 \(\displaystyle{P}_{{1}}{\left({1},{1}\right)},{P}_{{2}}{\left({1},−{1}\right)},{P}_{{3}}{\left(−{1},{1}\right)},{\quad\text{and}\quad}{P}_{{4}}{\left(−{1},−{1}\right)}\).

\(\displaystyle{{f}_{{x}}{\left({x},{y}\right)}}=\frac{\partial}{{\partial{x}}}{\left({x}^{{3}}+{y}^{{3}}-{3}{x}-{3}{y}+{1}\right)}\)

\(\displaystyle={3}{x}^{{2}}+{0}-{3}-{0}+{0}\)

\(\displaystyle={3}{x}^{{2}}-{3}\)

\(\displaystyle{{f}_{{y}}{\left({x},{y}\right)}}\frac{\partial}{{\partial{y}}}{\left({x}^{{3}}+{y}^{{3}}-{3}{x}-{3}{y}+{1}\right)}\)

\(\displaystyle={0}+{3}{y}^{{2}}-{0}-{3}+{0}\)

\(\displaystyle={3}{y}^{{2}}-{3}\)

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 \(\displaystyle{P}_{{1}}{\left({1},{1}\right)},{P}_{{2}}{\left({1},−{1}\right)},{P}_{{3}}{\left(−{1},{1}\right)},{\quad\text{and}\quad}{P}_{{4}}{\left(−{1},−{1}\right)}\).