Answer
in final $f$ has a saddle point at (0,0)
in final $f$ has a relative maximum at $\left(\frac{1}{6}, \frac{1}{12}\right)$
Work Step by Step
We are given that
\[
\begin{array}{l}
x y-x^{3}-y^{2}=f(x, y) \\
f_{x}(x, y)=x-2 y=y-3 x^{2}, \quad f_{y}(x, y)
\end{array}
\]
Consider $3 x^{2}-3 y=0$ and $3 y^{2}-3=0$
We get two critical points at (0,0) and $\left(\frac{1}{6}, \frac{1}{12}\right)$
By the theorem $13.8 .6,$ let $f$ be a function of two variables with continuous second-order partial derivatives in some disk centered at a critical point $\left(x_{0}, y_{0}\right),$ and let
\[
f_{x x}\left(x_{0}, y_{0}\right) f_{y y}\left(x_{0}, y_{0}\right)-f_{x y}^{2}\left(x_{0}, y_{0}\right)=D
\]
Do partial derivatives and check for a critical point at (0,0)
\[
\begin{array}{c}
f_{x x}(x, y)=-6 x, \quad f_{x y}(x, y)=1 \quad \text { and } \quad f_{y y}(x, y)=-2 \\
f_{x x}(0,0)=0, \quad f_{x y}(0,0)=1 \quad \text { and } \quad f_{y y}(0,0)=-2 \\
D=0 \times(-2)-(1)^{2} \\
D=-1<0
\end{array}
\]
Here $D<0,$ so $f$ has a saddle point at (0,0)
Check for a critical point at $\left(\frac{1}{6}, \frac{1}{12}\right)$
\[
\begin{array}{c}
f_{x x}(x, y)=-6 x, \quad f_{x y}(x, y)=1 \quad \text { and } \quad f_{y y}(x, y)=-2 \\
f_{x x}\left(\frac{1}{6}, \frac{1}{12}\right)=-1, \quad f_{x y}\left(\frac{1}{6}, \frac{1}{12}\right)=1 \quad \text { and } \quad f_{y y}\left(\frac{1}{6}, \frac{1}{12}\right)=-2 \\
(-1) \times(-2)-(1)^{2}=D \\
D=1>0
\end{array}
\]
Because here $D>0$ and $f_{x x}\left(\frac{1}{6}, \frac{1}{12}\right)<0,$ $f$ has a relative maximum at $\left(\frac{1}{6}, \frac{1}{12}\right)$
