Answer
There is a saddle point at $(\frac{3}{2},\frac{-11}{8})$
Work Step by Step
We are given $z=f(x,y)=x^{3}+2xy-4x-3y-2$
$f_{x}(x,y)=3x^{2}+2y-4$
$f_{y}(x,y)=2x-3$
Set the partial derivative equal to 0
$2x-3=0$
$x=\frac{3}{2}$
Now substitute for x in the other equation and solve for y
$3(\frac{3}{2})^{2}+2y-4=0$
$y=\frac{-11}{8}$
The critical point is $(\frac{3}{2},\frac{-11}{8})$
$f_{xx}(x,y)=6x$
$f_{yy}(x,y)=0$
$f_{xy}(x,y)=2$
$f_{xx}(\frac{3}{2},\frac{-11}{8}) =9$
$f_{yy}(\frac{3}{2},\frac{-11}{8})=0$
$f_{xy}(\frac{3}{2},\frac{-11}{8})=2$
$D=9\times0-(2)^{2}=-4$
Since D<0, there is a saddle point at $(\frac{3}{2},\frac{-11}{8})$
