Answer
Local minimum of $f(2,0)=\dfrac{1}{e^4}$
Work Step by Step
Since, we have $f_x(x,y)=(2x-4)e^{x^2+y^2-4x}=0, f_y(x,y)=2ye^{x^2+y^2-4x}=0$
After solving the above two equations, we get
critical point is: $(2,0)$
As per second derivative test, we have
$D=f_{xx}f_{yy}-f^2_{xy}=\dfrac{4}{e^8}$ and $D=\dfrac{4}{e^8} \gt 0$ and $f_{xx}=\dfrac{2}{e^4} \gt 0$
Thus, local minimum of $f(2,0)=\dfrac{1}{e^4}$
