Answer
Local minimum of $f(1,1)=3$
Work Step by Step
Since, we have $f_x(x,y)=\dfrac{-1}{x^2}=0, f_y(x,y)=x-\dfrac{1}{y^2}=0$
After solving the above two equations, we get
The critical point is: $(1,1$
As per the second derivative test, for $(1,1)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=3$ and $D=3 \gt 0$ and $f_{xx} =2 \gt 0$
Thus, we have: Local minimum of $f(1,1)=3$
