#### Answer

$(\pm \dfrac{1}{\sqrt 2},\dfrac{1}{2}) $ and $(\pm \dfrac{1}{\sqrt 2},-\dfrac{1}{2})$

#### Work Step by Step

Given: $f(x,y) =xy$
and $g(x,y)=x^2+2y^2-1=0$ ...(1)
Consider the gradient equation as follows: $\nabla =\lambda \nabla g$
Here $y=2 \lambda x $ or, $y=\pm \dfrac{1}{2 \sqrt 2}$
and $x=4 \lambda y $ or, $x=\pm \dfrac{2}{2 \sqrt 2}y$
Now, equation (1) becomes:
$g(x,y)=x^2+2y^2-1=0$
This implies that $x=\pm \dfrac{1}{\sqrt 2}$ and $y= \pm \dfrac{1}{2}$
Answer:
our points are: $(\pm \dfrac{1}{\sqrt 2},\dfrac{1}{2}) $ and $(\pm \dfrac{1}{\sqrt 2},-\dfrac{1}{2})$