Exercise 9.22

We have

so $D_{2}f(x,y)=0$ if $y=0$ or $x=-1∕2$. If $y=0$, then $D_{1}f(x,y)=6x(x-1)=0$ if $x=0$ or $x=1$. However, if $x=-1∕2$, then $D_{1}f(x,y)=6(3∕4+y^2)>0$ for all $y$. Hence the gradient of $f$ equals $\check{0}$ only at $(0,0)$ and $(1,0)$.

At $(1,0)$, let $d_1,d_2$ be near 0. Then

which is positive for small values of $d_1$ and $d_2$, hence $f$ has a local minimum at $(1,0)$.

At $(0,0)$, the values of $f$ along the $x$-axis, $f(x,0)=2x^3-3x^2$, have a local maximum at $x=0$, while the values of $f$ along $y$-axis, $f(0,y)=3y^2$ have a local minimum at $y=0$. Hence $f$ has a saddle point at $(0,0)$.

Solving $f(x,y)=0$ for $y$, we get

The graph of this looks like the folium of Descartes, only with a vertical asymptote of $x=-1∕2$ and symmetrical with the $x$-axis, with a double point at the origin and intersecting the $x$-axis at 0 and 3/2.

At the origin, a double-point, we cannot solve for $x$ in terms of $y$, or vice versa. Otherwise we can solve for $x$ in terms of $y$ except where $D_{2}f(x,y)=6y(2x+1)=0$, which on the zero set only occurs at the $x$-axis intercepts of the origin and $(3∕2,0)$. We can solve for $y$ in terms of $x$ away from the origin except where $D_{1}f(x)=6(x^2+y^2-x)=0$. Inserting the expression for $y$ above in this equation and solving for $x$, we see that this happens only at origin and the points