Answer
There is a saddle point at $(\frac{13}{3},\frac{-5}{3})$
Work Step by Step
We are given $z=f(x,y)=\frac{1}{2}x^{2}+\frac{1}{2}y^{2}+2xy-x-7y+10$
$f_{x}(x,y)=x+2y-1$
$f_{y}(x,y)=y+2x-7$
Set the partial derivative equal to 0
$x+2y-1=0$
$x=1-2y$
Now substitute for x in the other equation and solve for y
$y+2x-7=0$
$y+2(1-2y)-7=0$
$y=\frac{-5}{3} \rightarrow x=\frac{13}{3}$
The critical point is $(\frac{13}{3},\frac{-5}{3})$
$f_{xx}(x,y)=1$
$f_{yy}(x,y)=1$
$f_{xy}(x,y)=2$
$f_{xx}(\frac{13}{3},\frac{-5}{3}) =1$
$f_{yy}(\frac{13}{3},\frac{-5}{3})=1$
$f_{xy}(\frac{13}{3},\frac{-5}{3})=2$
$D=1\times1-(2)^{2}=-3$
Sinc D<0, there is a saddle point at $(\frac{13}{3},\frac{-5}{3})$
