Answer
$(\pm \dfrac{1}{\sqrt 2},\dfrac{1}{2}) $ and $(\pm \dfrac{1}{\sqrt 2},-\dfrac{1}{2})$
Work Step by Step
As we are given that $f(x,y) =xy$
and the equation of constraint $g(x,y)=x^2+2y^2-1=0$
since, we have the gradient equation $\nabla =\lambda \nabla g$
Now, $y=2 \lambda x \implies y=\pm \dfrac{1}{2 \sqrt 2}$
and $x=4 \lambda y \implies x=\pm \dfrac{2}{2 \sqrt 2}y$
Now,
$g(x,y)=x^2+2y^2-1=0 \implies y= \pm \dfrac{1}{2}; x=\pm \dfrac{1}{\sqrt 2}$
Hence, our points are: $(\pm \dfrac{1}{\sqrt 2},\dfrac{1}{2}) $ and $(\pm \dfrac{1}{\sqrt 2},-\dfrac{1}{2})$
