Answer
There is a saddle point at $(-2,\frac{5}{3})$
Work Step by Step
$f_{x}(x,y)=3y -5 = 0 \rightarrow y=\frac{5}{3}$
$f_{y}(x,y)=3x+6=0 \rightarrow x=-2$
The critical points is $(-2,\frac{5}{3})$
$f_{xx}(x,y)=0$
$f_{yy}(x,y)=0$
$f_{xy}(x,y)=3$
$D=0.0-3^{2}=-9$
Since there is D<0, there is a saddle point at $(-2,\frac{5}{3})$