Answer
f(x) has a relative minimum at $(-3,-3)$. This relative minimum is $f(-3,-3)=-19$
Work Step by Step
We are given $f(x,y)=3x^{2}-4xy+2y^{2}+6x-10$
$f_{x}(x,y)=6x -4y + 6 = 0$
$f_{y}(x,y)=-4x +4y=0 \rightarrow x=y$
Use the substitution method to solve the system of equations $6(y)-4y+6=0 \rightarrow y=-3$
$x=-3$
The critical point is $(-3,-3)$
$f_{xx}(x,y)=6$
$f_{yy}(x,y)=4$
$f_{xy}(x,y)=-4$
$D=6*4-(-4)^{2}=8$
Since $D\gt0$ and $f_{xx}(-3,-3)=6\gt0$ , f(x) has a relative minimum at $(-3,-3)$. This relative minimum is $f(-3,-3)=-19$
