I've done a bit: range of function: $[0,6]$. So we need level curves for $k = 0,1,\ldots,6$ $$f(x,y) = k$$ $$\sqrt{36 - 9x^2 - 4y^2} = k$$ $$36 - 9x^2 - 4y^2 = k^2$$ $$36-k^2 = 9x^2 + 4y^2$$
Not really sure how to go further. I tried to divide by $9$ and $4$ to simplify.. but it didn't work out.
$$36-k^2 = 9x^2+4y^2$$
$$1=\frac{x^2}{\left(\frac{\sqrt{36-k^2}}{3}\right)^2}+\frac{y^2}{\left( \frac{\sqrt{36-k^2}}{2}\right)^2}$$
Notice that $1=\frac{x^2}{a^2}+\frac{y^2}{b^2}$ is an equation of an ellipse.