Answer
Maximum: $f(\pm 1/\sqrt 2, \mp 1/(2\sqrt2))=e^{1/4}$ and Minimum: $f(\pm 1/\sqrt 2, \pm 1/(2\sqrt2))=e^{-1/4}$
Work Step by Step
Our aim is to calculate the extreme values with the help of the Lagrange Multipliers Method, subject to the given constraints. For this we have:$\nabla f(x,y)=\lambda \nabla g(x,y)$
This yields $\nabla f(x,y)=\lt -ye^{-xy},-xe^{-xy} \gt$ and $\lambda g(x,y)= \lt 2x,8y \gt$
Using the constraint condition $x^2+4y^2 \leq 1$ we get, $x^2= \dfrac{1}{2}$ and $x= \pm \dfrac{1}{\sqrt 2}$
Also, $\nabla f(x,y)=\lt -ye^{-xy},-xe^{-xy} \gt$ and $\lambda g(x,y)= \lt 2x,8y \gt$ we get $x=0,,y=0$
Hence, Maximum: $f(\pm 1/\sqrt 2, \mp 1/(2\sqrt2))=e^{1/4}$ and Minimum: $f(\pm 1/\sqrt 2, \pm 1/(2\sqrt2))=e^{-1/4}$
