For y I'm getting, y = \pm 1
The way I set this up was as follows:
Define:
f(x,y) = xy; g(x,y) = \frac{x^{2}}{9} + y^{2} = 2 \Rightarrow h(x,y,\lambda) = f(x,y) + \lambda g(x,y)
(Note that the sign in front of \lambda does not matter)
So let's take our partials, we get:
\frac{dh}{dx} = y + \frac{2 \lambda}{9}x, \frac{dh}{dy} = x + 2 \lambda y, \frac{dh}{d\lambda} = \frac{x^2}{9} +y^2 - 2
We know that each of those partials vanish i.e. we can set each to 0.
The first one gives us
9y = -2 \lambda x
and the second one gives us
\frac{x}{y} = -2 \lambda
And by simple substitution we get:
9y^{2} = x^{2}
So let's substitute it into our 3rd equation to get:
y^{2} + y^{2} = 2
Which yields our desired result of y = \pm 1. Now we can plug this into our g(x,y) to get x = \pm 3
Note that it doesn't matter which value for y we pick therefore our solution set will be:
(1,3), (1,-3), (-1,3), (-1,-3)
Now if you don't want to do this using Lagrange Multipliers, we can just realize that we can rewrite our g(x,y) as
y = \pm \sqrt{2 - \frac{x^2}{9}}
and now we can substitute this into our f(x,y) get an equation of one variable i.e.
\bar{f}(x) = \pm x \cdot \sqrt{2 - \frac{x^2}{9}}
Now we can proceed using the techniques you learned in Calculus 1 (I'm going to use Maple because I'm lazy)
> a:=x*sqrt(2-x^2/9);
a := \frac{1}{3} x \sqrt{18 - x}
> solve(diff(a,x)=0,x);
-3, 3
Note that choosing the negative root produces the same results.
