Calculus optimization problem Help I'm stuck on how to proceed with this problem.

Consider the function $f(x)=9-x^2$ for $f(x) \leq 0$. The shaded region is an isosceles triangle formed by joining the points $(0,0)$, $(x,f(x))$ and $(-x,f(x))$. What is the area of the largest triangle that satisfies the stated conditions (Look at picture)?

My teacher gave us a hint and I tried to do it as $A=1/2(3)(9-x^2)$. But I just have my doubts that I'm doing it wrong and I've looked everywhere for help and the only one who has it was Chegg, and I don't have access to that.
Here's how the figure should look like


Let the base of the triangle as $2x$ and the height as $f(x)$
$$A(x) = xf(x)$$
Set the derivative of $A(x)$ equal to zero
$$
\frac{d}{dx}A(x) = 0
$$
You'll found the $x$ points.
Use symmetry. Find the biggest triangle possible in the first quadrant and we know that the area of the full triangle is double it. The base of our triangle is given by $x$ and the height is given by $f(x)$ so we want to maximize $\frac{xf(x)}{2}=area$. 

$xf(x)=x(9-x^2)=9x-x^3$ take derivative to get $f’(x)=-3x^2+9$

and $x=\sqrt{3}$ makes it it zero which is the value that makes the area maximized.

By symmetry, the other point is at -$\sqrt{3}$