Chapter 13 - Section 13.7 - Extreme Values and Saddle Points - Exercises - Page 737: 22

Local minimum of $f(1,1)=3$

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 $f(1,1)=3$