Answer
There is a relative maximum at $(\frac{-3}{4},\frac{-9}{32})$. The relative maximum is $\frac{539}{128}$.
There is a saddle point at (0,0)
Work Step by Step
We are given $z=f(x,y)=x^{3}-8y^{2}+6xy+4$
$f_{x}(x,y)=3x^{2}+6y$
$f_{y}(x,y)=-16y+6x$
Set the partial derivative equal to 0
$-16y+6x=0$
$x=\frac{8}{3}y$
Now substitute for x in the other equation and solve for y
$3(\frac{8}{3}y)^{2}+6y=0$
$\frac{64}{3}y^{2}+6y=0$
$y=0 \rightarrow x=0$
or $y=\frac{-9}{32} \rightarrow x=\frac{-3}{4}$
There are two critical points $(0,0); (\frac{-3}{4},\frac{-9}{32})$
$f_{xx}(x,y)=6x$
$f_{yy}(x,y)=-16$
$f_{xy}(x,y)=6$
$f_{xx}(\frac{-3}{4},\frac{-9}{32}) =\frac{-9}{2}$
$f_{yy}(\frac{-3}{4},\frac{-9}{32})=-16$
$f_{xy}(\frac{-3}{4},\frac{-9}{32})=6$
$D=\frac{-9}{2}\times(-16)-(6)^{2}=36$
Since D>0, and $f_{xx}(\frac{-3}{4},\frac{-9}{32})=\frac{-9}{2}\lt0$, there is a relative maximum at $(\frac{-3}{4},\frac{-9}{32})$. The relative maximum is $\frac{539}{128}$
$f_{xx}(0,0) =0$
$f_{yy}(0,0)=-16$
$f_{xy}(0,0)=6$
$D=0\times(-16)-(6)^{2}=-36$
Since D<0, there is a saddle point at (0,0)
