So if we had the equation bound by a constant such that
$(L(x, y, λ) = 9x^2 - y^2 - λ(9x^2 + y^2 - 81)$
I can determine that
$\frac{∂L}{∂x} = 18x - 18λ$
$\frac{∂L}{∂y} = -2y - 2λ$
$\frac{∂L}{∂λ} = -(9x^2 + y^2 - 81)$
Assuming this is correct, when we equate each partial derivative to $0$ it may be clear that for $f_x, x = 1, λ = 1$ and presumably $y = 0$ but I have an issue with the constraint equation. I understand there are two answers and at first I assumed that $±9 - y = 9$ was the answer as $x = ±1$. However, I realised that there is a $-$ sign and wondered if that still made the application of the square root possible. Also, the $y$ is squared, so doesn't that mean that there is a $±y$ value as well? Is the value of $λ = 0$? Just not sure if I have equated the values correctly.
Use Lagrange Multiplier method for this problem only if you have not been given a choice. Now coming to your working, there is a mistake.
$L(x, y, λ) = 9x^2 - y^2 - λ(9x^2 + y^2 - 81)$. Equating partial derivatives to zero give you,
$18x - 18\lambda x = 0$
$-2y - 2\lambda y = 0$
$9x^2 + y^2 = 81$
From first, either $x = 0$ or $\lambda = 1 \,$ for $x \ne 0$.
If $\lambda = 1, y = 0$ from the second equation. So $x = \pm3$ given the constraint.
Similarly if $x = 0, y \ne 0,$ (due to constraint). Hence $\lambda = - 1 $ from second equation and $y = \pm 9$ given the constraint.
So you get critical points as $(\pm 3, 0), (0, \pm 9)$.